Answer
$ \displaystyle \frac{T_{1}}{T_{2}}=2^{1/4}$
Work Step by Step
The energy per time, or power $P$, radiated by an object with a surface area $A$ at the Kelvin temperature $T$ is
$P=e\sigma AT^{4} \qquad $16-17
where $e$ is the emissivity (a constant between $0$ and 1) and
$\sigma$ is the Stefan-Boltzmann constant, $\sigma=5.67\times 10^{-8}\mathrm{W}/(\mathrm{m}^{2}\cdot \mathrm{K}^{4})$.
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If $P_{1}=P_{2},\ A_{2}=A_{1}$, and $e_{2}=2e_{1}$,
$e_{1}\sigma A_{1}T_{1}^{4}=(2e_{1})\sigma A_{1}T_{2}^{4}$,
$\displaystyle \frac{T_{1}^{4}}{T_{2}^{4}}=2\quad \Rightarrow\quad \frac{T_{1}}{T_{2}}=2^{1/4}$
