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

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball into the tub, the height of the raised water is 6.75cm. Therefore, the volume of the raised water is

$V=\pi \times(12)^{2} \times 6.75 \mathrm{~cm}^{3}$

Let, the radius of the spherical ball is r. Therefore, the volume of the spherical ball is

$V_{1}=\frac{4}{3} \pi \times r^{3} \mathrm{~cm}^{3}$

Since, the volume of the raised water is same as the volume of the spherical ball, we have

$V_{1}=V$

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

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

$\Rightarrow \quad=12 \times 3 \times 6.75 \times 3$

$\Rightarrow \quad=9$