A conical container of base diameter $2R$ and height $H$ is full of water which is poured into a cylindrical container of diameter $2pR$. Then it will occupy a height equal to:

The base diameter of the conical container is $2R$, so its radius is $r_1=R$. Its height is $h_1=H$.
The volume of water in the conical container is given by:

$$V_{\text{cone}}=\frac{1}{3}\pi r_1^2{h}_1=\frac{1}{3}\pi R^{2}H$$

The water is poured into a cylindrical container of diameter $2pR$, so its radius is $r_2=pR$.
Let the height occupied by the water in the cylinder be $h_2$.
The volume of water in the cylindrical container is:

$$V_{\text{cylinder}}=\pi r_2^2{h}_2=\pi (pR{)}^2{h}_2=\pi p^2{R}^2{h}_2$$

Since the volume of water remains the same:

$$V_{\text{cylinder}}=V_{\text{cone}}$$

$$\pi p^2{R}^2{h}_2=\frac{1}{3}\pi R^{2}H$$

Canceling $\pi$ and $R^2$ from both sides:

$$p^2{h}_2=\frac{H}{3}$$

$$h_2=\frac{H}{3p^2}$$