= Solution
The equilibrium <photon gas> is <isotropic>, so the <energy density> per unit <solid angle> in a narrow angular-frequency interval is $u(T,\omega)d\omega/(4\pi)$. For rays making <angle> $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a <volume> $c\,dt\,dA\cos\theta$. Only the outward hemisphere contributes. Thus
$$
e(T,\omega)=\frac{cu(T,\omega)}{4\pi}\int_0^{2\pi}d\phi\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta
=\boxed{\frac{cu(T,\omega)}4}.
$$
The hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral <power> per unit area. A <blackbody> at the same <temperature> emits the same amount: otherwise radiation exchange between it and the aperture would violate <thermal equilibrium>.
Integrating <Planck's law> over all positive <angular frequencies> gives the total emitted <heat flux density>
$$
M(T)=\int_0^\infty e(T,\omega)\,d\omega
=\frac{\hbar}{4\pi^2c^2}\int_0^\infty\frac{\omega^3\,d\omega}{e^{\hbar\omega/(k_BT)}-1}.
$$
Set $x=\hbar\omega/(k_BT)$. Then
$$
M(T)=\frac{k_B^4T^4}{4\pi^2\hbar^3c^2}\int_0^\infty\frac{x^3\,dx}{e^x-1}.
$$
The remaining <integral> is independent of <temperature>, proving $M\propto T^4$. Its value follows by expanding $(e^x-1)^{-1}=\sum_{m=1}^\infty e^{-mx}$ and integrating termwise:
$$
\int_0^\infty\frac{x^3\,dx}{e^x-1}=\sum_{m=1}^\infty\frac6{m^4}=\frac{\pi^4}{15}.
$$
Consequently the <Stefan–Boltzmann law> is
$$
\boxed{M(T)=\sigma T^4,\qquad \sigma=\frac{\pi^2k_B^4}{60\hbar^3c^2}=5.67037\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}.}
$$
This is emitted <power> alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net <radiative cooling>.
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