The volume of a sphere is increasing at the rate of $4\pi cc/\sec$. Where  its volume is  $288\pi$cc,  the rate of increase (in cm/sec) in its radius is

Rate of increase $\frac{dv}{dx}=4\pi cc/\sec$
$v=288\pi cc$
As we know volume of a sphere of radius (r)
$v=\frac{4}{3}\pi r^3\mathrm{......}\left(1\right)$ 
Diff above expression w.r.t ‘t’ on both side we get
$\frac{dv}{dt}=\frac{d}{dt}\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\pi r^3\right)=\frac{4}{3}{\pi }^3{r}^2\frac{dr}{dt}$

$\begin{array}{l}\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}\\ eq\left(1\right)\left(288.\pi \right)=\frac{4}{3}\pi r^3\left(or\right)r=6cm\\ eq\left(2\right)weget\\ 4\pi =4\pi {\left(6\right)}^2\frac{dr}{dt}or\frac{dr}{dt}=\frac{1}{36}\end{array}$