Let points $O_1$, $O_2$, $O_3$, $O_4$ be the centers of the four solid iron balls respectively, with $O_1$, $O_2$ belonging to the two balls in the lower layer, and $A$, $B$, $C$, $D$ be the projective points of $O_1$, $O_2$, $O_3$, $O_4$ on the base of the cylinder. $ABCD$ constitute a square with the side of $\frac{\sqrt{2}}{2}$. So the height of the water in the cylinder must be $1 + \frac{\sqrt{2}}{2}$, so that all the balls are just immersed. Hence the volume of water we need is

$$\pi \left(1 + \frac{\sqrt{2}}{2}\right) - 4 \times \frac{4}{3} \pi \left(\frac{1}{2}\right)^3 = \left(\frac{1}{3} + \frac{\sqrt{2}}{2}\right) \pi.$$