= Solution
Because $T=T(\Psi)$,
$$
\nabla T=\frac{dT}{d\Psi}\nabla\Psi.
$$
The outward normal radiative flux is therefore
$$
\mathbf F\mathbin\cdot\widehat{\mathbf n}
=-\frac{4acT^3}{3\kappa\rho}
\frac{dT}{d\Psi}g.
$$
Integrating over the equipotential gives
$$
L_\Psi=-\frac{4acT^3}{3\kappa\rho}
\frac{dT}{d\Psi}S\langle g\rangle.
$$
Part (iii) gives $dm/d\Psi=\rho S\langle g^{-1}\rangle$, and hence
$$
\frac{dT}{d\Psi}=\rho S\langle g^{-1}\rangle\frac{dT}{dm}.
$$
Substitution yields
$$
\boxed{L_\Psi=-\frac{4acT^3}{3\kappa}
S^2\langle g\rangle\langle g^{-1}\rangle
\frac{dT}{dm}}.
$$
This is the radiative equation in the <equipotential stellar-structure approximation>.
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