A cylindrical tub of radius 12 cm contains water to a depth of 20 cm.

Radius of cylindrical tub = 12 cm

Depth = 20 cm

Let r be the radius of the ball

Then

Volume of the ball = Volume of water raised

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

$r^{3}=\frac{3.14 \times(12)^{2} \times 6.75 \times 3}{4}$

$\mathrm{r}^{3}=729$

$r=\sqrt[3]{729}$

r = 9 cm

Therefore radius of the ball = 9 cm