Rising water level

The volume of water in both pictures is $\pi*5^{2}*1.71 = 42.75\pi cm^{3}$ .

If we call the depth to be found $x$ , the second picture has the cone displacing water up to a frustum of height $x$ .

The volume of this frustum is $\frac{1}{3}\pi*4^{2}*8-\frac{1}{3}\pi*(4-\frac{1}{2}x)^{2}*(8-x) = (\frac{1}{12}x^{3}-2x^{2}-9x)\pi$

The volume of the water plus frustum must be equal to $25\pi x$

Setting the frustum as the difference between the two depth volumes gives the equation to be solved

$\frac{1}{12}x^{3}-2x^{2}-9x+42.75=0$

This cubic has extraneous solutions of $x=\frac{21\pm15\sqrt{5}}{2}$ . The solution of interest is exactly $x=\boxed{3cm}$