= Solution
Use the <single-layer greenhouse model> with a blackbody surface and one isothermal atmospheric layer. The layer is transparent to incoming stellar radiation, absorbs a fraction $\alpha$ of the surface's thermal radiation, and has the same thermal <emissivity> $\alpha$ by <Kirchhoff's law of thermal radiation>. Here $0\leq\alpha\leq1$, and $T_e$ is the <planetary equilibrium temperature> defined by the absorbed global stellar flux $\sigma_{\rm SB}T_e^4$. Neglect <convection>, latent heat transport and intrinsic heat.
The layer absorbs $\alpha\sigma_{\rm SB}T_s^4$ and emits $\alpha\sigma_{\rm SB}T_a^4$ upward and downward. For $\alpha>0$, its energy balance gives
$$
\alpha\sigma_{\rm SB}T_s^4=2\alpha\sigma_{\rm SB}T_a^4,\qquad T_a^4=T_s^4/2.
$$
The surface receives the absorbed stellar flux plus downward atmospheric emission, so
$$
\sigma_{\rm SB}T_s^4=\sigma_{\rm SB}T_e^4+\alpha\sigma_{\rm SB}T_a^4.
$$
Eliminating $T_a$ yields
$$
\boxed{T_s=\left(\frac{2}{2-\alpha}\right)^{1/4}T_e}.
$$
The outgoing top-of-atmosphere flux is $(1-\alpha)\sigma_{\rm SB}T_s^4+\alpha\sigma_{\rm SB}T_a^4=\sigma_{\rm SB}T_e^4$, confirming overall energy conservation. At $\alpha=0$, $T_s=T_e$ and the decoupled layer's <temperature> is not determined; at $\alpha=1$, $T_s=2^{1/4}T_e$. This warming is the restriction of thermal escape by the absorbing and emitting layer.
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