The volume of a sphere is increasing at 3 cubic centimeter per second.

Let r be the radius and V be the volume of the sphere at any time t. Then,

$V=\frac{4}{3} \pi r^{3}$

$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi r^{2}} \frac{d V}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{3}{4 \pi(2)^{2}} \quad\left[\because r=2 \mathrm{~cm}\right.$ and $\left.\frac{d V}{d t}=3 \mathrm{~cm}^{3} / \mathrm{sec}\right]$

$\Rightarrow \frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec}$