Firstly, I calculated the area of sector $AOB$ by applying $\frac{1}{2}\times (1.2\ \text{radians})\times 20^{2}$ (formula for area of sector of circle) and calculated area of sector $AOB= 240\ cm^2$, tried bit of construction by joining $PC$ and $CR$ thus reaching $CP \perp AO$ and $CR \perp CR$. Also, tried Joining $PR$ to form $\triangle POR$ but couldn't have the idea to find the value of '$x$'. Give it a try.
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In right $\triangle CPO$, we have $$\sin\frac{\angle AOB}{2}=\frac{PC}{OC}\quad (\text{since}\ \angle AOB=1.2\ \text{radians})$$ $$\implies \sin\frac{1.2}{2}=\frac{x}{20-x}$$ $$\implies x=20\sin 0.6-x\sin 0.6$$ $$\implies x=\frac{20\sin 0.6}{1+\sin 0.6}$$ $$\implies \color{blue}{x\approx 7.217527109\ cm}$$
Hint: $$\sin \angle COR = \frac{x}{20-x}$$