The minute hand of a clock is 15 cm long.

Angle inscribed by the minute hand in 60 minutes $=360^{\circ}$

Angle inscribed by the minute hand in 20 minutes $=\frac{360}{60} \times 20=120^{\circ}$

We have:

$\theta=120^{\circ}$ and $r=15 \mathrm{~cm}$

$\therefore$ Required area swept by the minute hand in 20 minutes $=$ Area of the sector with $r=15 \mathrm{~cm}$ and $\theta=120^{\circ}$

$=\frac{\pi r^{2} \theta}{360}$

$=3.14 \times 15 \times 15 \times \frac{120}{360}$

$=235.5 \mathrm{~cm}^{2}$