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 of radius 9cm into the tub, the height of the raised water is h cm. Therefore, the volume of the raised water is

$V=\pi \times(12)^{2} \times h$ cubic $\mathrm{cm}$

The volume of the spherical ball is

$V_{1}=\frac{4}{3} \pi \times(9)^{3}$ cubic $\mathrm{cm}$

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

$V_{1}=V$

$\Rightarrow \frac{4}{3} \pi \times(9)^{3}=\pi \times(12)^{2} \times h$

$\Rightarrow \quad h=\frac{4 \times(9)^{3}}{3 \times(12)^{2}}$

$\Rightarrow \quad=\frac{27}{4}$

$\Rightarrow \quad=6.75$

Therefore, the height of the raised water is $h=6.75 \mathrm{~cm}$