The relevant formula is the Stefan-Boltzman formula:

$\begin{align*}\frac{\Delta Q}{\Delta t} = \sigma \epsilon A T^4\end{align*}$
Here, $\Delta Q$ / $\Delta t$ is the rate at which energy is radiated from the blackbody, $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the emissivity of the blackbody (which is equal to 1 for an ideal Blackbody), $A$ is the area of the blackbody, and $T$ is the temperature in Kelvin.

Replacing $T$ by $3 T$ therefore increases the rate at which energy is radiated by a factor of $3^4 = 81$ . So, answer (E) is correct.

Solution to 2001 Problem 35