A solid copper cube and a solid copper sphere, both of same mass & emissivity are heated to same initial temperature and kept under identical conditions.  What is the ratio of their initial rate of fall of temperature?

If $\text{'}a\text{'}$ is the length of cube, surface area $S_{\text{cube}}=6a^2$ 
If $\text{'}r\text{'}$ is the radius of sphere, surface area  $S_{\text{sphere}}=4\text{π}{r}^2$
Both the cube and sphere have the same volume as given 
Volume  $V=a^3$  $\left(\because V=\frac{4}{3}\text{π}{r}^3\right)$
$\begin{array}{l}\therefore a^3=\frac{4}{3}\pi r^3\\ \Rightarrow \frac{a}{r}={\left(\frac{4\pi }{3}\right)}^{\frac{1}{3}}\\ \frac{{\text{ Area }}_{\text{cube }}}{{\text{ Area }}_{\text{sphere }}}=\frac{6a^2}{4\pi r^2}\\ =\frac{6}{4\pi }{\left(\frac{a}{r}\right)}^2=\frac{6}{4\pi }\times {\left(\frac{4\pi }{3}\right)}^{\frac{2}{3}}\\ ={\left(\frac{6}{\pi }\right)}^{\frac{1}{3}}\\ \text{∵ mass are equal }\end{array}$

Rate of fall in temperature = Surface area

$\because$Rate of fall in temperature= ${\left(\frac{6}{\text{π}}\right)}^{\frac{1}{3}}$