= Solution
Let $D\ll1$ be the remaining absorber <optical depth>, and retain the mixture's convective exponent $\kappa_a=R_a/c_{pa}$. The outgoing surface contribution is $\sigma T_s^4e^{-D}\simeq\sigma T_s^4(1-D)$. To first order, atmospheric emission is the column-mean source times $D$. If $4\kappa_a\simeq1$, the <adiabat> gives $T(p)^4\simeq T_s^4p/p_a$, a source linear in mass/optical depth. Therefore its mean is $T_s^4/2$, or equivalently it is evaluated at $p_a/2$:
$$
F_{\rm out}\simeq\sigma T_s^4(1-D)+D\sigma T_s^4/2
=\sigma T_s^4(1-D/2).
$$
Setting this equal to $L_{\circledast}/4$ gives
$$
\boxed{T_s\simeq\left[\frac{L_{\circledast}}{2\sigma(2-\tau_\infty)}\right]^{1/4}.}
$$
The zero-opacity limit correctly gives $T_e$, while a small positive opacity raises the surface <temperature> slightly. The expression is a first-order thin-column result, not a formula to extend to its apparent pole at $\tau_\infty=2$. More generally, direct thin-column averaging gives $F_{\rm atm}\simeq D\sigma T_s^4/(1+4\kappa_a)$; the half-column value is exact at this order when $4\kappa_a=1$. The assumed convective thermal structure remains a modelling condition even when the gas becomes radiatively thin.
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