Boltzmann built on two known properties of cavity radiation.
(1) The energy density, $u$, defined as $u = U/V$, depends only on temperature, $T$.
(2) The radiation pressure, $p$ is given by $p= u/3$.
Radiation pressure was given a firm basis c1862 by Maxwell. The factor of 1/3 arises because of the three-dimensionality of the cavity, in which radiation is propagating in all possible directions. [It's easy for us now to derive this equation by considering the cavity as containing a photon gas.]
Boltzmann (1884) used a thought-experiment in which a cavity is fitted with a piston, and we take the radiation inside it through a Carnot cycle. On a $p–V$ diagram the isothermals are just horizontal lines, because $u$ is constant so $p$ is constant. The heat input along the top (temperature $T$) isothermal is $\Delta U + p \Delta V$. This works out to be $4 p \Delta V$. If the lower temperature isothermal is only slightly lower, at temperature ($T - d T$) then the cycle appears as a thin horizontal box, and the net work done during the cycle is simply $dp \Delta V$ in which $dp$ is the infinitesimal pressure difference between the two isothermals. We can then apply the definition of thermodynamic temperature in the form
$$\frac{dT}{T} = \frac{\text{work}}{\text{heat input}} = \frac{dp \Delta V}{4 p \Delta V} = \frac{dp}{4 p}~.$$
This integrates up easily to give the Stefan-Boltzmann law.
A modern thermodynamic derivation would probably start from the fundamental equation (embodying first and second laws of thermodynamics)… $$\text {d}U = T \text{d}S - P \text{d}V. $$ Therefore for an isothermal change $$\left(\frac{\partial U}{\partial V} \right)_T = T \left(\frac{\partial S}{\partial V} \right)_T - p .$$ Using one of the Maxwell relations, this becomes $$\left(\frac{\partial U}{\partial V} \right)_T = T \left(\frac{\partial p}{\partial T} \right)_V - p .$$ But for the radiation, $$\left(\frac{\partial U}{\partial V} \right)_T = \left(\frac{\partial (uV)}{\partial V} \right)_T = u = 3p \ \ \ \ \ \text{and} \ \ \ \ \ \left(\frac{\partial p}{\partial T} \right)_V = \frac {\text{d} p}{\text{d}T}. $$ Making these substitutions and tidying, $$ 4p = T \frac {\text{d} p}{\text{d}T}.$$ As before, separation of variables and integration gives the Stefan-Boltzmann law.
The solution to your problem(1) is this :
\begin{align}
R(T) & =\frac c4 \int\limits_{\lambda=0}^{\lambda=\infty} \lambda^{-5}f(\lambda T)\mathrm{d}\lambda=\frac c4 \int\limits_{\lambda=0}^{\lambda=\infty} T^{4}\dfrac{f(\lambda T)}{(\lambda T)^{5}}\mathrm{d}(\lambda T)\qquad \Longrightarrow
\nonumber\\
R(T) & =\frac c4 \underbrace{\left(\:\:\int\limits_{\mu=0}^{\mu=\infty} \dfrac{f(\mu)}{\mu^{5}}\mathrm{d}\mu\right)}_{A=\text{constant}}T^{4}=
\underbrace{\left(\frac c4 A\right)}_{\sigma}\, T^{4}=\sigma\, T^{4}
\tag{01}
\end{align}
But, sincerely, trying to find this directly you are missing important facts about the "before Planck" adventure of the blackbody radiation theory. For example, you must try to find from Wien's Law (your first equation) why if you know the function $\;\rho(\lambda,T_{1})\;$ for a given temperature $\;T_{1}\;$ then you know it for any temperature $\;T\;$ or that $\;\lambda_{\rm max}\cdot T=b=\rm constant\;$ (Wien's Displacement Law), see Emilio Pisanty answer therein : Showing Wien's Displacement Law from Wien's Law.
(1)
"Quantum Mechanics" B. H. Bransden-C. J. Joachain, 2nd Edition 2000, Pearson Education Limited (Problem 1.3, page 45)
Best Answer
Wien's fifth power law applies to the maximum height of the emissive power density. The Stefan–Boltzmann law applies to the total emissive power (the integration of the emissive power density).
From Wien's displacement law (derived here), $\lambda_\max T = b$ is constant. So, from Planck's law (e.g. see here):
$I'(\lambda_\max, T) \propto \frac{\lambda_\max^{-5}}{e^{hc/(\lambda_\max k T)} -1} = \frac{T^5/b^5}{e^{hc/(k b)} -1} \propto T^5$
and integrating Plank's law (e.g. here except I keep in terms of wavelength for slightly longer) we get, by changing variables to $x=hc/(\lambda kT)$ so that $\lambda^{-5}d\lambda \propto -T^4x^3dx$ (the negative is absorbed by changing the integration interval limits):
$\int_0^\infty I'(\lambda,T)d\lambda \propto \int_0^\infty \frac{T^4 x^3dx}{e^{x}-1} \propto T^4$.