= Solution
{c}
In <local thermodynamic equilibrium>, write the one-ray <radiative transfer equation> as
$$
I_\nu+\kappa_\nu^{-1}\frac{dI_\nu}{dz}=B_\nu[T(z)].
$$
Large total <optical depth> removes the boundary contribution, but an additional local condition is essential: the <temperature> and <opacity> vary little over an absorption length $\kappa_\nu^{-1}$. Iterating once in that small gradient gives
$$
I_\nu=B_\nu-\frac1{\kappa_\nu}\frac{\partial B_\nu}{\partial T}\frac{dT}{dz}+\text{higher gradients}.
$$
The integrated <Planck function> and its derivative are
$$
\int_0^\infty B_\nu\,d\nu=\frac{\sigma T^4}{\pi},\qquad
\int_0^\infty\frac{\partial B_\nu}{\partial T}\,d\nu=\frac{4\sigma T^3}{\pi}.
$$
Using the specified <Rosseland mean opacity>, with $\kappa_R$ measured here in inverse length, the source's definition $F_{\rm ray}=\int I_\nu d\nu$ therefore gives
$$
\boxed{F_{\rm ray}(z)=\frac{\sigma T^4}{\pi}-\frac{4\sigma T^3}{\pi\kappa_R}\frac{dT}{dz}+\text{higher gradients}.}
$$
The printed leading term $\sigma T^4$ lacks a factor $1/\pi$ under that definition. An isothermal optically thick medium is a concrete counterexample: $I_\nu=B_\nu$ gives $F_{\rm ray}=\sigma T^4/\pi$ exactly, while the printed formula gives $\sigma T^4$.
A physical <radiative flux> requires an angular integral. For direction cosine $\mu$, the same gradient expansion gives $I_\nu(\mu)=B_\nu-\mu\kappa_\nu^{-1}\partial_zB_\nu$. Integration over all directions gives the net <radiative diffusion> flux
$$
\boxed{F_{\rm net}=2\pi\int_{-1}^1\mu\,d\mu\int_0^\infty I_\nu(\mu)\,d\nu
=-\frac{16\sigma T^3}{3\kappa_R}\frac{dT}{dz}.}
$$
The isotropic equilibrium contribution cancels between opposite directions. For comparison the outward-hemisphere flux alone is $\sigma T^4-8\sigma T^3(dT/dz)/(3\kappa_R)$ at this order. None of these consistent definitions yields the printed mixture of coefficients.
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