= Solution
Use the <Einstein coefficients> convention in which $B_{ij}$ multiplies spectral <radiation energy density> $u_\nu$, with $u_\nu=c^{-1}\int I_\nu\,d\Omega$. A normalized <spectral line profile> obeys $\int\varphi(\nu)d\nu=1$. The <spontaneous emission> probability per upper-state atom per unit time is $A_{21}$; the <stimulated emission> and upward <absorption> rates per atom are respectively $B_{21}\int u_\nu\varphi(\nu)d\nu$ and $B_{12}\int u_\nu\varphi(\nu)d\nu$.
For <isotropic> <spontaneous emission>, $n_2A_{21}$ transitions occur per unit volume and time, each supplying <photon energy> $h\nu_{12}$. Dividing among $4\pi$ steradians gives
$$
\boxed{j_\nu=\frac{h\nu_{12}}{4\pi}n_2A_{21}\varphi(\nu).}
$$
For a small incident beam, its contribution to $u_\nu$ is $I_\nu d\Omega/c$. <Absorption> removes and <stimulated emission> adds <photons> in the incident direction. Matching the net loss to $\alpha_\nu I_\nu d\Omega$ therefore gives
$$
\boxed{\alpha_\nu=\frac{h\nu_{12}}{c}(n_1B_{12}-n_2B_{21})\varphi(\nu).}
$$
Replacing $\nu_{12}$ by $\nu$ across the narrow <spectral line profile> gives the printed expressions. The negative sign is essential: an inverted population can amplify the incident <specific intensity>. <Thermal equilibrium> and the <Planck function> imply $g_1B_{12}=g_2B_{21}$ and $A_{21}/B_{21}=8\pi h\nu_{12}^3/c^3$, where $g_i$ is the <state degeneracy>. A convention using <mean intensity> instead changes both $B$ coefficients by $4\pi/c$.
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