= Solution
The incident stellar power is intercepted over a disk and distributed over the spherical surface. With a black shortwave surface, no <clouds> and no shortwave atmospheric absorption, the global mean absorbed flux is $F_*=L_{\circledast}/4$. Define $T_e=(F_*/\sigma)^{1/4}$.
To relate a radiating-level <temperature> to the ground, assume the intervening well-mixed gas follows a <adiabatic process>[dry adiabat] with constant $R_g,c_{pg}$:
$$
T(p)=T_s\left(\frac p{p_s}\right)^{\kappa_g},\qquad\kappa_g=\frac{R_g}{c_{pg}}.
$$
Emission from $p_{\rm rad}$ balances absorbed sunlight when $\sigma T(p_{\rm rad})^4=F_*$. Hence
$$
\boxed{T_s=\left(\frac{L_{\circledast}}{4\sigma}\right)^{1/4}
\left(\frac{p_s}{p_{\rm rad}}\right)^{R_g/c_{pg}}
\simeq T_e\tau_\infty^{R_g/c_{pg}}.}
$$
The final form uses the optically thick result in part (a). Since $p_{\rm rad}<p_s$, the surface is warmer than the effective emission <temperature>. Efficient horizontal redistribution alone does not mathematically fix the vertical lapse rate: the adiabatic/convective assumption is needed to obtain the requested dependence on $R_g/c_{pg}$. A different imposed vertical profile would give a different relation.
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