= Solution
The <Stefan–Boltzmann law> gives solar <luminosity> $L_\odot=4\pi R_\odot^2\sigma T_\odot^4$. At orbital distance $r$, the incident <irradiance> is $\sigma T_\odot^4R_\odot^2/r^2$. A spherical absorber intercepts a disc of <area> $\pi R_M^2$ but radiates over $4\pi R_M^2$. Thus the uniform-temperature <radiative equilibrium> estimate is
$$
\pi R_M^2\sigma T_\odot^4\frac{R_\odot^2}{r^2}=4\pi R_M^2\sigma T_M^4,
\qquad
\boxed{T_M=T_\odot\sqrt{\frac{R_\odot}{2r}}\simeq245\,\mathrm K.}
$$
The planetary <radius> cancels. This is a <black-body radiation> estimate with zero reflected fraction and complete redistribution of absorbed <heat>. More generally the balance fixes $\langle T^4\rangle^{1/4}$, rather than the arithmetic mean of a nonuniform surface <temperature>. It excludes atmospheric heating and thermal lag; no measured surface <temperature> is inferred.
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