Given that
area of cuboid = 160 cm2
Level of water is increased = 2 cm
Volume of vessel = 160 × 2 cm3 ……. (1)
Volume of each sphere = \(\frac{4}{3}π R^3\)
Volume of 3 sphere = \(3\times{\frac{4}{3}}πR^3\) ……. (2)
From eq 1 and eq 2
⇒ 3 × (\(\frac{4}3\)) π R3 = 160 × 2
⇒ \(R^3 = \frac{160\times2}{3\times\frac{4}{3}π}\)
⇒ R = 2.94 cm