Step 1: Here $r=\dfrac{21}{2}=10.5$ cm and $\theta=120^\circ$.
Step 2: Length of arc ACB:
$$\ell=\frac{120}{360}\times 2\times\frac{22}{7}\times \frac{21}{2}=\frac13\times 66=22\ \text{cm}$$
Step 3: Length of chord AB. Drop $OM\perp AB$, then $\angle AOM=60^\circ$:
$$AB=2r\sin 60^\circ=2\times 10.5\times\frac{\sqrt3}{2}=10.5\sqrt3$$
Step 4: $AB=10.5\times 1.7=17.85$ cm
Step 5: Perimeter of segment ACB $=$ arc ACB $+$ chord AB
$$=22+17.85=39.85\ \text{cm}$$
Answer: $39.85$ cm