Let the radius of the sphere is ‘r’ and the edge of the cube is ‘a’ and let the volume of both be 1 cubic units, then
$\dfrac{4}{3}\pi r^3 = 1$ and $a^3 = 1$
⇒ $ r = \left(\dfrac{3}{4\pi}\right)^{1/3}$ and $a = 1$
Let the surface areas of sphere and cube be ‘S’ and ‘C’, then
$\dfrac{S}{C} = \dfrac{4\pi r^2}{ 6a^2}$
=$4\pi \times \left(\dfrac{3}{4\pi}\right)^{2/3}: 6$
=$4\pi \times \left(\dfrac{3}{4\pi}\right)^{2/3}: 6$
