If the radius of the spherical ball is increasing at the rate of 1 m/sec and the radius of the ball is 3 metre than the volume of the ball is increasing at the rate of

$36 \pi \, m^3/sec$

Given $\frac{dr}{dt}=1m/sec$, r → radius of ball

$v=\frac{4}{3}\pi r^3$, v → volume of ball

$\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$

so $\frac{dv}{dt}=4\pi (3)^2×1=36 \pi \, m^3/sec$