Given,

Surface area of cube $=$ Surface area of sphere

$\Rightarrow 6 a^2 =4 \pi r^2 \dots(i)$

$\Rightarrow \frac{a}{r}=\sqrt{\left(\frac{4 \pi}{6}\right)}=\sqrt{\frac{2 \pi}{3}}$

Now, buoyant force is $F_B=V_{\text {in }} \cdot \rho_f . g$ So, ratio of buoyant force on cube and sphere is

$\frac{\left(F_B\right)_{\text {cube }}}{\left(F_B\right)_{\text {sphere }}}=\frac{V_{\text {cube }}}{V_{\text {sphere }}}$

$=\frac{a^3}{\frac{4}{3} \pi r^3}=\frac{3}{4 \pi} \times\left(\frac{2 \pi}{3}\right)^{\frac{3}{2}}$

$=\sqrt{\frac{9}{16 \times \pi^2} \times 27}=\sqrt{\frac{\pi}{6}}$

$\therefore \left(F_B\right)_{\text {cube }} < \left(F_B\right)_{\text {sphere }}$