The radius of a circle is increasing uniformly at the rate of 3 cm/s.

The area of a circle (A) with radius (r) is given by,

$A=\pi r^{2}$

Now, the rate of change of area (A) with respect to time (t) is given by,

$\frac{d A}{d t}=\frac{d}{d t}\left(\pi r^{2}\right) \cdot \frac{d r}{d t}=2 \pi r \frac{d r}{d t}$   [By chain rule]

It is given that,

$\frac{d r}{d t}=3 \mathrm{~cm} / \mathrm{s}$

$\therefore \frac{d A}{d t}=2 \pi r(3)=6 \pi r$

Thus, when r = 10 cm,

$\frac{d A}{d t}=6 \pi(10)=60 \pi \mathrm{cm}^{2} / \mathrm{s}$

Hence, the rate at which the area of the circle is increasing when the radius is $10 \mathrm{~cm}$ is $60 \pi \mathrm{cm}^{2} / \mathrm{s}$.