The twenty-three

Consider the figure on the left. It is a frustum of a pyramid and the volume is given by $V=\dfrac{h}{3}(A_1+A_2+\sqrt{A_1 \times A_2})$ where $A_1$ and $A_2$ are the base areas and $h$ is the height. By similar triangles, we have

$\dfrac{x}{2}=\dfrac{2}{3}$ $\implies$ $x=\dfrac{4}{3}$

$A_1=2^2=4$ and $A_2=\left(\dfrac{4}{3}\right)^2=\dfrac{16}{9}$

So the volume of water is $V=\dfrac{1}{3}\left(4+\dfrac{16}{9}+\sqrt{4\left(\dfrac{16}{9}\right)}\right)=\dfrac{76}{27}$

Consider the figure on the right. The volume of water is given by $V=\dfrac{1}{3}y^2h$ where $y^2$ is the base area and $h$ is the height of water. By similar triangles, we have

$\dfrac{y}{h}=\dfrac{2}{3}$ $\implies$ $y=\dfrac{2}{3}h$

Substitute,

$\dfrac{76}{27}=\dfrac{1}{3}\left(\dfrac{2}{3}h\right)^2h$ $\implies$ $\dfrac{2052}{108}=h^3$ $\implies$ $h^3=19$ $\implies$ $h=\boxed{\sqrt[3]{19}\qquad}$