Answer
$\lambda = 2.42~\mu m$
Work Step by Step
We can find the temperature of the cube:
$P = e~\sigma~A~T^4$
$T^4 = \frac{P}{e~\sigma~A}$
$T = \sqrt[4] {\frac{P}{e~\sigma~A}}$
$T = \sqrt[4] {\frac{630~W}{(1)~(5.67\times 10^{-8}~W~m^{-2}~K^{-4})~(6)~(3.0\times 10^{-2}~m)^2}}$
$T = \sqrt[4] {2.057613169\times 10^{12}~K^4}$
$T = 1197.7~K$
We can use Wien's law to find the peak wavelength:
$\lambda = \frac{2.9\times 10^6~nm~K}{T}$
$\lambda = \frac{2.9\times 10^6~nm~K}{1197.7~K}$
$\lambda = 2.42\times 10^{3}~nm$
$\lambda = 2.42\times 10^{-6}~m$
$\lambda = 2.42~\mu m$
