Step 1 :Given that the volume of the water is increasing at a rate of $3 m^3/min$, we need to find how fast the height of the water is increasing. The volume of a cylinder is given by $V = \pi r^2 h$. Differentiating both sides with respect to time $t$, we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Step 2 :Substituting the given values into the equation, we get $3 = \pi (4)^2 \frac{dh}{dt}$. Solving for $\frac{dh}{dt}$, we get $\frac{dh}{dt} = \frac{3}{16\pi}$.
Step 3 :Thus, the height of the water is increasing at a rate of $\boxed{\frac{3}{16\pi}}$ meters per minute.