A spherical ball of radius 3 cm is melted and recast into three spherical balls.

Radius of sphere = 3 cm
Radius of first ball = 1.5 cm
Radius of second ball = 2 cm
Let radius of the third ball be r cm.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls

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

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

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

$=36 \pi-\left(\pi \frac{9}{2}+\frac{32}{3} \pi\right)$

$=\left(\frac{36 \times 6-9 \times 3-32 \times 2}{6}\right) \pi$

$=\left(\frac{216-27-64}{6}\right) \pi=\frac{125 \pi}{6}$

Therefore,

$\frac{4}{3} \pi r^{3}=\frac{125 \pi}{6}$

$\mathrm{O} r, \mathrm{r}=\sqrt[3]{\frac{125 \times 3}{4 \times 6}}=\sqrt[3]{\frac{125}{8}}=\frac{5}{2}=2.5 \mathrm{~cm}$