A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm × 22 cm × 14 cm).

Volume of the rectangular solid of iron = 32 cm × 22 cm × 14 cm 

Radius of the container, $r=\frac{56}{2}=28 \mathrm{~cm}$

Let the rise in the level of water in the container when rectangular solid of iron is submerged in it be h cm.

$\therefore$ Volume of the water displaced in the container $=\pi r^{2} h=\pi \times(28)^{2} \times h$

When the rectangular solid of iron is submerged in the container, then the volume of water displaced in the container is equal to the volume of the rectangular solid of iron.

$\therefore \frac{22}{7} \times(28)^{2} \times h=32 \times 22 \times 14$

$\Rightarrow h=\frac{32 \times 22 \times 14 \times 7}{22 \times(28)^{2}}$

$\Rightarrow h=4 \mathrm{~cm}$

Thus, the rise in the level of water in the container is 4 cm.