Inspired by Suprem.S.Nalkund

According to the problem, $6{ l }^{ 2 }=4\pi { r }^{ 2 }$

${ l }^{ 2 }=\frac { 4\pi { r }^{ 2 } }{ 6 }$

${ l }=\sqrt { \frac { 4\pi { r }^{ 2 } }{ 6 } }$

therfore the volume of the cube is $={ \left( \sqrt { \frac { 4\pi { r }^{ 2 } }{ 6 } } \right) }^{ 3 }$

$=\frac { 4\pi { r }^{ 2 } }{ 6 } \times 2r\sqrt { \frac { \pi }{ 6 } }$

$=\frac { 4\pi { r }^{ 3 } }{ 3 } \sqrt { \frac { \pi }{ 6 } }$

and the volume of the sphere is, $=\frac { 4\pi { r }^{ 3 } }{ 3 }$

Therefore the ratio is, $\frac { \frac { 4\pi { r }^{ 3 } }{ 3 } }{ \frac { 4\pi { r }^{ 3 } }{ 3 } .\sqrt { \frac { \pi }{ 6 } } }$

which is nothing but, $\frac { 1 }{ \sqrt { \frac { \pi }{ 6 } } }$

i.e, $\sqrt { \frac { 6 }{ \pi } }$

which is approximately, 1.381.

Let the surface areas of the two solids be $100$ .

For the sphere:

$100=4 \pi r^2$ $\implies$ $r^2=\dfrac{25}{\pi}$ $\implies$ $r=\dfrac{5}{\sqrt{\pi}}$

$V=\dfrac{4}{3} \pi \left(\dfrac{5}{\sqrt{\pi}}\right)^3=94.032$

For the cube:

$100=6a^2$ $\implies$ $a^2=\dfrac{50}{3}$ $\implies$ $a=\sqrt{\dfrac{50}{3}}$

$V=\left(\sqrt{\dfrac{50}{3}}\right)^3=68.04$

$Ratio=\dfrac{94.032}{68.04}=1.382$

nice solution