Consider a smooth spherical shell. The outside of the shell admits no energy transfer via radiation or any other means, so that the shell and its interior form an isolated system. The shell is at temperature $T$. Its inside surface is a perfect black body.
Next put a smooth solid sphere, also a perfect black body, in the center of the shell, also at temperature $T$. It will absorb radiation from the inside surface of the shell and will also emit radiation, all of which is absorbed by the shell. Because the shell and ball are at the same temperature, they cannot exchange net heat, so the sphere absorbs and emits radiation at the same rate.
Next replace the sphere with a perfect black body golf ball of equal cross section to the sphere. This ball absorbs the same amount of radiation from the shell as the sphere did simply because it absorbs all the radiation that hits it. The golf ball, like the solid sphere, must emit and absorb the same amount of radiation because otherwise heat would flow between bodies of equal temperature. Therefore, the golf ball emits the same amount of radiation as the sphere.
If you start with a sphere and put dimples in it, you've actually reduced the cross-section a bit because the caps of the dimples are gone, so such a dimpled sphere would radiate slightly less than before.
The Stefan-Boltzmann law is
$$
\frac{P}{A}=\frac{2\pi^5 k^4}{15 h^3 c^2} T^4,
\tag{1}
$$
where $P/A$ denotes the emitted power per unit surface area of the blackbody. In equation (1), the coefficient of $T^4$ is the Stefan-Boltzmann constant. In contrast, the energy per unit volume ($E/V$) of the radiation inside the blackbody is
$$
\frac{E}{V}=\frac{8\pi^5 k^4}{15 h^3 c^3} T^4.
\tag{2}
$$
In equation (2), the coefficient of $T^4$ is the radiation constant. Equation (2) is used, for example, to compute the energy density of the cosmic microwave background radiation, because in that case we are inside the blackbody (the universe). These two quantities are related to each other by
$$
\frac{P}{A} = \frac{c}{4}\times \frac{E}{V}.
$$
The question is, where does the factor of $4$ come from?
The essence of the answer is that inside the blackbody, the radiation at any point is coming and going equally in all directions; but outside the blackbody, that is no longer true. The distribution over directions is not uniform outside the body, and the difference ends up being a factor of $4$. Equation (1) describes the emitted power outside the blackbody, whereas equation (2) describes the energy density inside.
The derivation of both equations, (1) and (2), involves the quantity $B(\lambda)$ that was given in the OP. Consider the quantity
$$
\frac{4\pi}{c}B(\lambda)\,d\lambda
\tag{3}
$$
where $d\lambda$ is an infinitesimal range of wavelengths. The quantity (3) is the energy density of the electromagnetic (EM) radiation that is contained inside the blackbody, within the given range of wavelengths. The factor of $4\pi$ accounts for the full sphere's worth of possible directions of the radiation at each point inside the blackbody. Integrating (3) over all wavelengths gives the total energy density (2).
To derive the Stefan-Boltzmann law (1), suppose for a moment that radiation is only able to escape through a small hole of area $A$ in the surface of the blackbody. How much energy leaks out per unit time? If we consider plane waves moving in a particular direction, then the amount that escapes through the hole depends on the direction in which the plane wave is moving compared to the orientation of the hole. So, to calculate this, we should start with the quantity
$$
\frac{1}{c}B(\lambda)\,d\lambda,
\tag{4}
$$
which is the energy density per unit solid angle inside the blackbody. This is obtained from (3) by omitting the factor of $4\pi$. EM radiation travels at speed $c$, so the energy per unit time that escapes through the hole into a given narrow cone of solid angle $d\Omega$ is
$$
\frac{1}{c}B(\lambda)\,d\lambda\times cA\cos\theta\,d\Omega
\tag{4}
$$
where $\theta$ is the angle of the radiation's direction of travel relative to the direction normal to the hole. The factor $A\cos\theta$ is the area of the hole projected orthogonally to the direction in which the radiation is traveling. This is the key. To calculate the rate at which energy escapes, we should integrate the quantity (4) over all directions with $\theta < \pi/2$. Directions with $\theta > \pi/2$ don't contribute, because that radiation is moving toward the inside of the body instead of toward the outside. Doing this integral (and cancelling the factors of $c$) gives
$$
B(\lambda)\,d\lambda\times A
\int_0^{2\pi}d\phi\int_0^{\pi/2} d\theta\,\sin\theta\,\cos\theta
= B(\lambda)\,d\lambda\times A\pi,
\tag{5}
$$
so the quantity
$$
\pi B(\lambda)\,d\lambda
\tag{6}
$$
is the power per unit surface area of the radiation that escapes from the hole. Even though this analysis considered radiation escaping through a "hole" in the surface, the same analysis applies to every piece of the surface of a blackbody that allows radiation to escape. The Stefan-Boltzmann law comes from integrating (6) over all wavelengths.
Best Answer
I think this is just matter of geometry related to the fact that the surface volume of a sphere is four times the apparent cross-section. Imagine yourself between two parallel plates. First consider the energy from one plate. If it radiating parallel beams of light perpendicular to its surface, the energy density would just be P/c. But the "average speed" of energy coming off the surface is not really the speed of light, because the average angle is not 90 degrees. This gives a factor of 2...there's twice as much energy density as the "naive" calculation. But this still isn't the energy density of the black body field, because you're looking at the vaccuum of empty space on one side. That's why you need the second plate. This doubles the energy density again, giving you the factor of four.
EDIT: Okay, I've redone this as carefully as I can, and I'm not getting the factor of four.
I start with a hollow sphere of radius 1 meter and let the temperature be such that the radiated power is 1 watt/m^2. I need to figure out how much energy is inside that sphere. (The naive calculation says it's just volume*power/c.) But this doesn't really work because the energy is criss-crossing in all directions.
I get around this by allowing the shell to instantaneously disintegrate so the energy explodes outwards in a spherical shell. At a sufficiently great distance, the energy is all flowing in the same direction so the naive calculation becomes correct. Letting c=1, this is where I find the energy to be located:
The factor of pi/4 on the shell energy comes from the profile of the energy pulse...it actually replicates the cross section of the sphere. When I put the total shell energy back into the volume of the original sphere, I get an energy density of 3*pi/2. The calculation looks right to me but it doesn't line up with either of your sources.
EDIT 2: Okay, I found my mistake! I did some 2-d integration that should have been in 3 dimensions. This diagram explains my corrections (shown in green):
The power which shows up in a given time interval delta-t is the energy which came from the disc as shown. The areas of these discs are proportional to an inverted parabola (area factor 2/3) instead of a half-circle (area factor pi/4). When I make this correction the energy density in the spherical volume comes out to 4, as hoped for.
(I can't explain an additional factor of pi that seems to show up in your Wikipedia reference.)