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 $5cm$