The edge of a cube is equal to the radius of the sphere. If the rate at which the volume of the cube is increasing is equal to '$\lambda$., then the rate of increase of volume of the sphere, is  

Let r be the radius of the sphere. Let $V_1$and $V_2$ be 

the volumes of the sphere and the cube whose edge is r. Then,

$V_1=\frac{4}{3}\pi r^3$and $V_2=r^3$

$\Rightarrow \frac{d{V}_1}{dt}=4\pi r^2\frac{dr}{dt}$ and $\frac{d{V}_2}{dt}=3r^2\frac{dr}{dt}$

$\Rightarrow \frac{d{V}_1}{dt}=4\pi r^2\frac{dr}{dt}$ and $\lambda =3r^2\frac{dr}{dt}$

$\Rightarrow \frac{d{V}_1}{dt}=\frac{4\pi \lambda }{3}$