Answer
-6.1 nW
Work Step by Step
Recall: The net rate of energy exchange
$P_{net}=\sigma\varepsilon A(T_{env}^{4}-T^{4})$
Given/known: Stefan-Boltzmann constant $\sigma=5.6704\times10^{-8}\,W/m^{2}\cdot K^{4}$,
Emissivity $\varepsilon=0.75$,
Surface area $A=6(6.0\times10^{-6}\,m)^{2}=2.16\times10^{-10}\,m^{2}$
Surface temperature(in K) $T=(-100+273)K=173\,K$
Temperature of the environment(in K) $T_{env}=(-150+273)K=123\,K$
Result:
$P_{net}=5.6704\times10^{-8}\,W/m^{2}\cdot K^{4}\times0.75\times2.16\times10^{-10}\,m^{2}((123\,K)^{4}-(173\,K)^{4})=-6.1\times10^{-9}\,W=-6.1\,nW$
