A hemispherical bowl of internal radius 9 cm is full of water.

We have,

the radius of the hemispherical bowl, $R=9 \mathrm{~cm}$ and

the internal base radius of the cylindrical vessel, $r=6 \mathrm{~cm}$

Let the height of the water in the cylindrical vessel be $h$.

As,

Volume of water in the cylindrical vessel $=$ Volume of hemispherical bowl

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

$\Rightarrow r^{2} h=\frac{2}{3} R^{3}$

$\Rightarrow 6 \times 6 \times h=\frac{2}{3} \times 9 \times 9 \times 9$

$\Rightarrow h=\frac{2}{3} \times \frac{9 \times 9 \times 9}{6 \times 6}$

$\Rightarrow h=\frac{27}{2}$

$\therefore h=13.5 \mathrm{~cm}$

So, the height of the water in the cylindrical vessel is 13.5 cm.