The radius of a sphere is changing. At an instant of time the rate of change in its volume and its surface area are equal. Then the value of radius at that instant is?

Given that, at any instant of time 
Rate of change in volume w.r.t. time = rate of change in surface area w.r.t. time
i.e.,    $\frac{dv}{dt}=\frac{ds}{dt}\mathrm{............}\left(i\right)$
Volume of sphere of radius  $\left(r\right)$ ,  $V=\frac{4}{3}\pi r^3$
Differentiating w.r.t, ‘t’, we get
 $$\begin{gathered} \frac{dv}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi \frac{d}{dt}\left(r^3\right) \\ \frac{dv}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}or\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}\mathrm{..........}\left(ii\right) \end{gathered}$$
Surface area of sphere of radius (r),
 $S=4\pi r^2$
Differentiating w.r.t. ‘t’, we get
 $\frac{ds}{dt}=\frac{d}{dt}\left(4\pi r^2\right)=4\pi \frac{d}{dt}\left(r^2\right)$
 $\Rightarrow \frac{ds}{dt}=8\pi r\frac{dr}{dt}\mathrm{...............}\left(iii\right)$
Putting the values from equation (ii) and (iii) in equation (i), we get
 $4\pi r^2\frac{dr}{dt}=8\pi r\frac{dr}{dt}orr=2$