A conical flask is full of water.

The base-radius and height of the conical flask are r and h respectively. Let, the slant height of the conical flask is l. Therefore, the volume of the water in the conical flask is

$V=\frac{1}{3} \times \pi \times r^{2} \times h$

The water in the conical flask is poured into a cylindrical flask of base-radius mr. Let, the height of the water in the cylindrical flaks is h1. Then, the volume of the water in the cylindrical flaks is

$V_{1}=\pi \times(m r)^{2} \times h_{1}$

Since, the volume of the water in the cylindrical flaks is same as the volume of the water in the conical flaks, we have

$V_{1}=V$

$\Rightarrow \pi \times(m r)^{2} \times h_{1}=\frac{1}{3} \times \pi \times r^{2} \times h$

$\Rightarrow \quad m^{2} \times h_{1}=\frac{1}{3} \times h$

$\Rightarrow \quad h_{1}=\frac{h}{3 m^{2}}$

Therefore, the height of the water in the cylinder is $\frac{h}{3 m^{2}}$