A ball floats in water in a prism-shaped glass. If we remove the ball, by how much will the water level in the glass sink, if the glass has a regular hexagonal base, each side of which is $5 \mathrm{~cm}$, the floating ball has a radius of $4 \mathrm{~cm}$, and the part of the ball protruding above the water is $1 \mathrm{~cm}$ high?
If the radius of the sphere is $r$, and the height of the submerged slice is $m$, then the volume of the slice is

$$
K=\frac{\pi m^{2}}{3}(3 r-m)
$$

If we take the sphere out of the water, the water level in the glass will drop by $x \mathrm{~cm}$ as a result. Then:

$$
x \cdot \frac{3 \sqrt{3} \cdot a^{2}}{2}=\frac{\pi m^{2}}{3}(3 r-m)
$$

where $a$ is one edge of the prism's base. Substituting the given values, $x=3.95 \mathrm{~cm}$.