A vessel in the shape of a cuboid contains some water.

The area of the base of the cuboid is $160 \mathrm{~cm}^{2}$. After immersing three identical spheres the level of the water is increased by $2 \mathrm{~cm}$. Therefore, the volume of the increased water is

$V=160 \times 2 \mathrm{~cm}^{3}$

Let the radius of each of the spheres is r cm. Then, the volume of each of the sphere is

$V_{1}=\frac{4}{3} \pi \times(r)^{3} \mathrm{~cm}^{3}$

The total volume of the three spheres is

$V_{2}=3 \times \frac{4}{3} \pi \times(r)^{3}=4 \pi \times(r)^{3} \mathrm{~cm}^{3}$

Since, the volume of the increased water is equal to the total volume of the three spheres; we have

$V_{2}=V$

$\Rightarrow 4 \pi \times(r)^{3}=160 \times 2$

$\Rightarrow \quad r^{3}=\frac{320 \times 7}{4 \times 22}$

$\Rightarrow \quad r^{3}=25.45$

$\Rightarrow \quad r=2.94$