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.,
$\frac{d v}{d t}=\frac{d s}{d t}$ $\ldots(i)$
Volume of sphere of radius $(r), V=\frac{4}{3} \pi r^3$
Differentiating w.r.t, ' $t$ ', we get
$\frac{d v}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi \frac{d}{d t}\left(r^3\right)$
$\frac{d v}{d t}=\frac{4}{3} \pi\left(3 r^2\right) \frac{d r}{d t}$ or $\frac{d v}{d t}=4 \pi r^2 \frac{d r}{d t}$ $\ldots(ii)$
Surface area of sphere of radius $(r)$,
$S=4 \pi r^2$
Differentiating w.r.t. ' $t$ ', we get
$\frac{d s}{d t}=\frac{d}{d t}\left(4 \pi r^2\right)=4 \pi \frac{d}{d t}\left(r^2\right) \Rightarrow \frac{d s}{d t}=8 \pi r \frac{d r}{d t} \ldots$ (iii)
Putting the values from Eqs. (ii) and (iii) in Eq. (i), we get
$4 \pi r^2 \frac{d r}{d t}=8 \pi r \frac{d r}{d t} \text { or } r=2$