Rate of emission of energy $=\sigma {\text{T}}^4\text{S}$

Let m₁ be the mass of sphere, C is specific heat and  $\left(\text{d}\theta /\text{dt}\right)$, the rate of cooling.

For sphere

$\sigma {\text{T}}^4\text{S}={\text{m}}_1\text{C}{\left(\frac{\text{d}\theta }{\text{dt}}\right)}_{\text{S}}$            ...(i)

Let m₂ be the mass of cube, C its specific heat and $\left(\text{d}\theta /\text{dt}\right)$, the rate of cooling

For cube

$\sigma {\text{T}}^4\text{S}={\text{m}}_2\text{C}{\left(\frac{\text{d}\theta }{\text{dt}}\right)}_{\text{C}}$            ...(ii)

From Eqs.(i) and (ii)

         $\frac{{\left(\text{d}\theta /\text{dt}\right)}_{\text{s}}}{{\left(\text{d}\theta /\text{dt}\right)}_{\text{c}}}=\frac{{\text{m}}_2}{{\text{m}}_1}$
$\mathrm{=}\frac{{\text{a}}^3\rho }{\left(4/3\right)\pi {\text{r}}^2\rho }$

where a is the side of cube and r is the radius of sphere, $\rho$ is the density.

Required ratio $\frac{{\text{R}}_{\text{s}}}{{\text{R}}_{\text{c}}}=\frac{3{\text{a}}^3}{4\pi {\text{r}}^3}$

But since $\text{S }$(surface area) is the same,

      $6{\text{a}}^2=4\pi {\text{r}}^2$

or  ${\text{a}}^2=\left(2/3\right)\pi {\text{r}}^2$

$\therefore \frac{{\text{R}}_{\text{s}}}{{\text{R}}_{\text{c}}}=\frac{3{\left(2\pi {\text{r}}^2/3\right)}^{3/2}}{4\pi {\text{r}}^3}=\frac{2\pi \sqrt{2\pi }}{\sqrt{3}\left(4\pi \right)}$

           $=\sqrt{\frac{2\pi }{12}}=\sqrt{\frac{\pi }{6}}$