Answer
$$\approx 2.3 \times 10^{2} \mathrm{W} .
$$
Work Step by Step
Half the body surface area is roughly $A=1.8 / 2=0.9 \mathrm{m}^{2} . $
Now, with $T_{\mathrm{env}}=248 \mathrm{K},$ we find
$$
\left|P_{\text {net }}\right|=\left|\sigma \varepsilon A\left(T_{\text {env }}^{4}-T^{4}\right)\right| \approx 2.3 \times 10^{2} \mathrm{W} .
$$
