= Solution
{c}
Use the energy-density convention of part (a), a normalized common <spectral line profile> $\int\phi(\nu)\,d\nu=1$, and isotropic <spontaneous emission>. For a narrow <spectral line> centered at $\nu_0$,
$$
j_\nu=\frac{h\nu_0}{4\pi}n_2A_{21}\phi(\nu),\qquad
\kappa_\nu=\frac{h\nu_0}{c}(n_1B_{12}-n_2B_{21})\phi(\nu).
$$
The factor $1/c$ follows from $u_\nu=4\pi J_\nu/c$: angle-integrated absorption removes $c\kappa_\nu u_\nu$ of <energy> per volume and time. The <emission coefficient> is per <solid angle>, whereas the <absorption coefficient> has units of inverse length.
The <Boltzmann distribution> in <thermal equilibrium> gives $n_2/n_1=(g_2/g_1)e^{-h\nu_0/(k_BT)}$. Dividing the two expressions and using the relations between <Einstein coefficients>,
$$
\begin{aligned}
S_\nu&=\frac{c}{4\pi}\frac{n_2A_{21}}{n_1B_{12}-n_2B_{21}}\\
&=\frac{2h\nu_0^3/c^2}{e^{h\nu_0/(k_BT)}-1}
=B_{\nu_0}(T).
\end{aligned}
$$
Thus the <source function> equals the <Planck function> at the transition <frequency>, and in the usual narrow-line approximation
$$
\boxed{S_\nu=B_\nu(T).}
$$
With an exactly common finite-width $\phi$ and constant line-center <Einstein coefficients>, their ratio is constant across the profile; exact equality to a varying $B_\nu$ at every frequency is only a narrow-line approximation. Exact thermal <radiative transfer> requires frequency-resolved <detailed balance>, which supplies the corresponding small profile corrections. This distinction does not alter the intended equilibrium result.
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