= Solution
Let the spacecraft radius be $a$, and assume that <heat> redistribution keeps its whole surface at a common <temperature> $T$. Neglect background radiation, and let $R\gg r,a$. The Sun's <luminosity>, from the <Stefan–Boltzmann law>, is
$$
L_S=4\pi r^2\sigma T_S^4.
$$
At distance $R$, the incident <irradiance> is $F_S=L_S/(4\pi R^2)=\sigma T_S^4(r/R)^2$. The spacecraft intercepts radiation over its projected disc, area $\pi a^2$, but emits from its whole spherical area $4\pi a^2$. The <radiative equilibrium> condition is therefore
$$
\pi a^2\sigma T_S^4\frac{r^2}{R^2}=4\pi a^2\sigma T^4.
$$
Cancelling the common factors gives
$$
\boxed{T=T_S\sqrt{\frac r{2R}}\ \propto\ T_S\sqrt{\frac rR}.}
$$
The radius of the spacecraft cancels. The factor $1/\sqrt2$ follows from the ratio of absorbing projected area to emitting surface area. Uniform surface <temperature> is the intended additional idealization behind a single equilibrium surface <temperature>. A poorly conducting spacecraft would have position-dependent <temperatures>; for example, a locally reradiating patch facing the Sun would instead have $T_{\rm subsolar}=T_S\sqrt{r/R}$, and a different maximum-temperature constraint.
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