The temperature and the surface property determine the power per unit area; multiply by the emitting area:

$$P=e\sigma A{T}^4=0.75(5.67\times {10}^{-8})\left(0.80\right){\left(320\right)}^4$$

$P=3.57\times {10}^2\mathrm{W}$.

This is the outgoing emission. We have not yet calculated absorption from the surroundings.

Divide by the same emitting area:

$$I_{\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{d}}=P\mathrm{/}A=356.7\mathrm{/}0.80$$

$I_{\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{d}}=4.46\times {10}^2\mathrm{W}{\mathrm{m}}^{-2}$.