= Solution
The given formula is the integral of the <Planck photon distribution>. Set
$$
x=\frac{h\nu}{k_BT},
\qquad d\nu=\frac{k_BT}{h}\,dx.
$$
Then
$$
n=\frac{8\pi}{c^3}\left(\frac{k_B}{h}\right)^3T^3
\int_0^\infty\frac{x^2}{e^x-1}\,dx
=\alpha T^3,
$$
where
$$
\alpha=\frac{8\pi}{c^3}\left(\frac{k_B}{h}\right)^3
\int_0^\infty\frac{x^2}{e^x-1}\,dx.
$$
Each photon of frequency $\nu$ has energy $h\nu$, so the energy density is
$$
\begin{aligned}
\rho
&=\frac{8\pi h}{c^3}\int_0^\infty
\frac{\nu^3}{e^{h\nu/(k_BT)}-1}\,d\nu\\
&=\frac{8\pi k_B^4}{c^3h^3}T^4
\int_0^\infty\frac{x^3}{e^x-1}\,dx
=\xi T^4.
\end{aligned}
$$
This is the <temperature scaling of thermal photon number and energy densities>.
After decoupling, <cosmological redshift> gives
$$
\nu(t)=\frac{a(t_{\rm dec})}{a(t)}\nu_{\rm dec},
\qquad
\nu_{\rm dec}=\frac{a(t)}{a(t_{\rm dec})}\nu(t).
$$
The occupation number is conserved along the freely propagating photon trajectory. Its exponential argument at decoupling becomes
$$
\frac{h\nu_{\rm dec}}{k_BT_{\rm dec}}
=\frac{h\nu(t)}{k_B\bigl(a(t_{\rm dec})T_{\rm dec}/a(t)\bigr)}.
$$
Thus the spectrum still has the Planck form if
$$
\boxed{T(t)=\frac{a(t_{\rm dec})}{a(t)}T_{\rm dec}}.
$$
Frequency and temperature redshift by the same factor, which is the <redshift preservation of a thermal photon spectrum>.
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