Answer
$\frac{dh}{dt} = \frac{3}{25\pi} \frac{m}{min}$
Work Step by Step
Substitute the $r$ value and differentiate the equation to the volume of a cylinder.
$V = \pi r^{2}h$
$V = \pi 5^{2}h$
$V = 25\pi h$
$\frac{dV}{dt} = 25\pi \frac{dh}{dt}$
Now substitute $\frac{dV}{dt}$ and solve it for $\frac{dh}{dt}$.
$ 3 = 25\pi \frac{dh}{dt}$
Now solve for $\frac{dh}{dt}$:
$\frac{dh}{dt} = \frac{3}{25\pi} \frac{m}{min}$
