= Solution
Each independent cavity mode is a <quantum harmonic oscillator> of <angular frequency> $\omega$. Its occupation numbers $n=0,1,\ldots$ represent the numbers of <photons> in that mode. At <thermal equilibrium>, <photons> can be absorbed and emitted by the walls, so their number is not conserved and the <photon chemical potential> is zero. The <Boltzmann distribution> of the oscillator excitations then gives
$$
\langle n\rangle=\frac1{e^{\hbar\omega/(k_BT)}-1},\qquad
\bar U=\hbar\omega\langle n\rangle.
$$
The expression supplied in the paper is the thermal excitation <energy>. The oscillator's temperature-independent <zero-point energy> is excluded from the emitted <thermal radiation>.
Putting $\epsilon=\hbar\omega$ and $d\epsilon=\hbar\,d\omega$ into the mode count from (a) gives the number of modes per unit cavity <volume>:
$$
\frac{dN}{L^3}=\frac{\omega^2}{\pi^2c^3}\,d\omega.
$$
Multiplying by the mean thermal <energy> in each mode yields <Planck's law> in the angular-frequency convention:
$$
\boxed{u(T,\omega)\,d\omega=\frac{\omega^2\,d\omega}{\pi^2c^3}\frac{\hbar\omega}{e^{\hbar\omega/(k_BT)}-1}.}
$$
The two <photon polarizations> are already included in the <density of states>; no further factor of two is needed. $u(T,\omega)d\omega$ has units $\mathrm{J\,m^{-3}}$, whereas $u(T,\omega)$ has units $\mathrm{J\,s\,m^{-3}}$.
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