The Wikipedia article on "Stellar density" says the stellar density near the Sun is only 0.14 stars per cubic parsec. It suggests that the density in the central core and in globular clusters is about 500 times as great.
According to the List of nearest stars and brown dwarfs in Wikipedia , there are 61 stars within 15 light years of the Earth. Dividing 61 by the volume of sphere of this radius, we obtain 4.3e-3 stars per cubic light-year, or 0.15 stars per cubic parsec. Thanks to user31264 for providing this information, which is consistent with the information from the previous link.
According to the Wikipedia article on "apparent magnitude", the total integrated magnitude of the night sky as seen from Earth is -6.5. Making that 500 times as bright produces a total magnitude of about -13.2 (5 magnitudes is a factor of 100 in brightness). The maximum brightness of the full Moon is -12.92.
So even with 500 times as many stars in the sky, the total brightness would be only slightly greater than that of a full moon.
(This assumes that the average brightness of the core stars is similar to the average brightness out here in the Galactic suburbs.)
Parts of the core might be even denser than that.
(I've updated this with new information and deleted and old link whose numbers appear to have been incorrect.)
It turns out that it is the distribution of birth stellar masses and most importantly, the lifetimes of stars as a function of mass that are responsible for your result.
Let's fix the number of stars at 200 billion. Then let's assume they follow the "Salpeter birth mass function" so that $n(M) \propto M^{-2.3}$ (where $M$ is in solar masses) for $M>0.1$ to much larger masses. There are more complicated mass function known now - Kroupa multiple power laws, Chabrier lognormal, which say there are fewer low mass stars than predicted by Salpeter, but they don't change the gist of the argument.
Using the total number of stars in the Galaxy, we equate to the integral of $N(M)$ to get the constant of proportionality: thus
$$n(M) = 1.3\times10^{10} M^{-2.3}.$$
Now let's assume most stars are on the main sequence and that the luminosity scales roughly as $L = M^{3.5}$ ($L$ is also in solar units), thus $dL/dM = 3.5 M^{2.5}$.
We now say $n(L) = n(M)\times dM/dL$ and obtain
$$ n(L) = 3.7\times10^{9} M^{-4.8} = 3.7\times10^{9} L^{-1.37}.$$
The total luminosity of a collection of star between two luminosity intervals is
$$ L_{\rm galaxy} = \int^{L_2}_{L_1} n(L) L \ dL = 5.9\times 10^{9} \left[L^{0.63} \right]^{L_{2}}_{L_1}$$
This equation shows that although there are far more low-mass stars than high mass stars in the Galaxy, it is the higher mass stars that dominate the luminosity.
If we take $L_1=0.1^{3.5}$ we can ask what is the upper limit $L_2$ that gives $L_{\rm galaxy} = 1.3\times 10^{10} L_{\odot}$ ($=5\times10^{36}$ W)?
The answer is only $3.5L_{\odot}$. But we see many stars in the Galaxy that are way brighter than this, so surely the Galaxy ought to be much brighter?
The flaw in the above chain of reasoning is that the Salpeter mass function represents the birth mass function, and not the present-day mass function.
Most of the stars present in the Galaxy were born about 10-12 billion years ago.
The lifetime of a star on the main sequence is roughly $10^{10} M/L = 10^{10} M^{-2.5}$ years. So most of the high mass stars in the calculation I did above have vanished long ago, so the mass function effectively begins to be truncated above about $0.9M_{\odot}$. But that also then means that because the luminosity is dominated by the most luminous stars, the luminosity of the galaxy is effectively the number of $\sim 1M_{\odot}$ stars times a solar luminosity.
My Salpeter mass function above coincidentally does give that there are $\sim 10^{10}$ star with $M>1M_{\odot}$ in the Galaxy. However you should think of this as there have been $\sim 10^{10}$ stars with $M>1 M_{\odot}$ born in our Galaxy. A large fraction of these are not around today, and that is actually the lesson one learns from the integrated luminosity number you quote!
EDIT: A postscript on some of the assumptions made. The Galaxy is much more complicated than this. "Most of the stars present in the Galaxy were born 10-12 billion years go". This is probably not quite correct, depending on where you look. The bulge of the Galaxy contains about 50 billion stars and was created in the first billion years or so. The halo also formed early and quickly, but probably only contains a few percent of the stellar mass. The moderately metal-poor thick disk contains perhaps another 10-20% and was formed in the first few billion years. The rest (50%) of the mass is in the disk and was formed quasi-continuously over abut 8-10 billion years. (Source - Wyse (2009)). None of this detail alters the main argument, but lowers the fraction of $>1M_{\odot}$ stars that have been born but already died.
A second point though is assuming that the luminosity of the Galaxy is dominated by main sequence stars. This is only true at ultraviolet and blue wavelengths. At red and infrared wavelengths evolved red giants are dominant. The way this alters the argument is that some fraction of the "dead" massive stars are actually red giants which typically survive for only a few percent of their main sequence lifetime, but are orders of magnitude more luminous during this period. This means the contribution of of the typical low-mass main sequence stars that dominate the stellar numbers is even less significant than the calculation above suggests.
Best Answer
The first picture is a view of the center of the galaxy, as observed from Earth. There's quite a lot of dust in between it and Earth, so on many wavelengths (including visible light), we can't see much.
The first picture is actually only part of a larger picture, including the Paranal Observatory:
Image courtesy of Wikipedia user Nikthestunned, under the Creative Commons Attribution 3.0 Unported license.
The galactic enter corresponds to the big, bright bulge in the second picture. This holds Sagittarius A*, a radio source within the region Sagittarius A that is thought to be a supermassive black hole.
The Milky Way is a spiral galaxy (specifically, a barred spiral). It has four arms, although the discovery of the "New Outer Arm" has cast doubt on whether or not there may be a fifth.
We're in the plane of the Milky Way, so we can't view it face-on and see its spiral structure in the same way we can view other galaxies. That's why photos of it - your second image is an artist's impression, not a photo - are typically bar-shaped. However, using a fish-eye lens can cause it to appear quite curved, as in this photograph, also from Paranal:
Image courtesy of Wikipedia user Soerfm, under the Creative Commons Attribution 3.0 Unported license.
As a final note, I must add that I'm somewhat confused by the use of "bulge" in the second picture. In this context, "bulge" refers to the central group of stars in the galaxy. The leader in the second picture, however, appears to be pointing to the end of one of the spiral arms, which is inaccurate.