**Step 1 — Write the given data.**
$$r = 21 \text{ cm}, \qquad \theta = 120^\circ, \qquad \pi = \frac{22}{7}$$

**Step 2 — Area of sector OAYB.**
$$= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21$$
$$= \frac13 \times \frac{22}{7} \times 441 = \frac13 \times 22 \times 63 = \frac{1386}{3} = 462 \text{ cm}^2$$

**Step 3 — Area of $\triangle$ OAB.**
Using $\text{Area} = \dfrac12 r^2 \sin\theta$ :
$$= \frac12 \times 21 \times 21 \times \sin 120^\circ = \frac12 \times 441 \times \frac{\sqrt3}{2}$$
$$= \frac{441\sqrt3}{4} = 110{\cdot}25\sqrt3 \approx 190{\cdot}95 \text{ cm}^2$$

**Step 4 — Area of segment = Area of sector – Area of triangle.**
$$= 462 - \frac{441\sqrt3}{4}$$

**Step 5 — Numerical value (taking $\sqrt3 = 1{\cdot}73$).**
$$\approx 462 - 190{\cdot}73 = 271{\cdot}27 \text{ cm}^2$$

$$\boxed{\text{Area of segment AYB} = \left(462 - \frac{441\sqrt3}{4}\right) \text{cm}^2 \approx 271{\cdot}3 \text{ cm}^2}$$