= Solution
For an ideal <blackbody> surface completely surrounded by an isothermal environment, the <Stefan–Boltzmann law> gives emitted <power> $A\sigma T_s^4$ and absorbed <power> $A\sigma T_e^4$. The net outward <radiative heat transfer> is therefore
$$
\boxed{P_{\rm net}=A\sigma(T_s^4-T_e^4)=2(5.670374419\times10^{-8})(308.15^4-293.15^4)\simeq185\,\mathrm W.}
$$
The gross emitted <power> is about $1022\,\mathrm W$, but about $837\,\mathrm W$ is absorbed back from the surroundings. The difference, rather than the emitted term alone, is the relevant <energy> loss.
\Image[/past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2021-skin-radiation.png]
{title=Skin and surroundings emit overlapping thermal spectra; their difference gives the net radiative loss}
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