A cylindrical tub of radius 16 cm contains water to a depth of 30 cm.

Let r be the radius of the iron ball

Radius of the cylinder = 16 cm

Then,

Volume of iron ball = Volume of water raised in the hub

$\frac{4}{3} \pi r^{3}=\pi r^{2} h$

$\frac{4}{3} r^{3}=(16)^{2} \times 9$

$\mathrm{r}^{3}=\frac{27 \times 16 \times 16}{4}$

$r^{3}=1728$

r = 12 cm

Therefore radius of the ball = 12 cm.