A drinking glass is in the shape of the frustum of a cone of height 21 cm

Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h.

Then, $R \Rightarrow \frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm}, r \Rightarrow \frac{4}{2} \mathrm{~cm}=2 \mathrm{~cm}, h=21 \mathrm{~cm}$

Capacity of the glass = Capacity of the frustum of the cone

$=\frac{\pi h}{3}\left[R^{2}+r^{2}+R r\right]$

$=\frac{22}{7} \times \frac{1}{3} \times 21 \times\left[(3)^{2}+(2)^{2}+(3 \times 2)\right] \mathrm{cm}^{3}$

$=(22 \times 19) \mathrm{cm}^{3}$

$=418 \mathrm{~cm}^{3}$