A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm.

We have,

height of cone, $h=60 \mathrm{~cm}$,

the base radius of cone, $r=30 \mathrm{~cm}$,

the height of cylinder, $H=180 \mathrm{~cm}$ and

the base radius of the cylinder, $R=60 \mathrm{~cm}$

Now,

Volume of water left in the cylinder = Volume of cylinder-Volume of cone

$=\pi R^{2} H-\frac{1}{3} \pi r^{2} h$

$=\frac{22}{7} \times 60 \times 60 \times 180-\frac{1}{3} \times \frac{22}{7} \times 30 \times 30 \times 60$

$=\frac{22}{7} \times 30 \times 30 \times 60\left(2 \times 2 \times 3-\frac{1}{3}\right)$

$=\frac{22}{7} \times 54000\left(12-\frac{1}{3}\right)$

$=\frac{22}{7} \times 54000 \times \frac{35}{3}$

$=1980000 \mathrm{~cm}^{3}$

$=\frac{1980000}{1000000} \mathrm{~m}^{3}$

$=1.98 \mathrm{~m}^{3}$

So, the volume of water left in the cylinder is 1.98 m3.