According to the question it is given,
Radius of the spherical ball $=3cm$
As we know that,
The volume of the sphere $=4/3\pi {{r}^{2}}$
Now, it’s volume (V) $=4/3\pi {{r}^{3}}$
That the ball is melted and recast into 3 spherical balls.
Volume $\left( {{V}_{1}} \right)$ of first ball $=4/3\pi {{1.5}^{3}}$
Volume $\left( {{V}_{2}} \right)$ of second ball $=4/3\pi {{2}^{3}}$
Assume the radius of the third ball $=rcm$
Volume of third ball $\left( {{V}_{3}} \right)=4/3\pi {{r}^{3}}$
Volume of the spherical ball is equal to the volume of the 3 small spherical balls.
$V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}$
$\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{1.5}^{3}}+\frac{4}{3}\pi {{2}^{3}}+\frac{4}{3}\pi {{r}^{3}}$
Then,
Cancelling out the common part from both sides of the equation we get,
${{\left( 3 \right)}^{3}}={{\left( 2 \right)}^{3}}+{{\left( 1.5 \right)}^{3}}+{{r}^{3}}$
${{r}^{3}}={{3}^{3}}-{{2}^{3}}-{{1.5}^{3}}c{{m}^{3}}$
${{r}^{3}}=15.6c{{m}^{3}}$
$r={{\left( 15.6 \right)}^{1/3}}cm$
$r=2.5cm$
As diameter $=2\times radius=2\times 2.5cm$
$=5.0cm$
Thus, the diameter of the third ball is $5cm$