$\begin{align}  & r=\frac{d}{2}=2\sqrt{3}cm \\ & \text{Area of sector }=\frac{\theta }{{{360}^{\text{o}}}}\times \pi {{r}^{2}} \\ & =\frac{{{120}^{\text{o}}}}{{{360}^{\text{o}}}}\times \pi {{(2\sqrt{3})}^{2}} \\ & =\frac{{{120}^{\text{o}}}}{{{360}^{\text{o}}}}\times 12\pi =4\pi c{{m}^{2}} \\ & \text{Area of circle}=\pi {{r}^{2}}= \\ & =\pi {{(2\sqrt{3})}^{2}}=12\pi c{{m}^{2}} \\ & \text{Area of the major sector }=(12\pi -4\pi )c{{m}^{2}}=8\pi c{{m}^{2}} \end{align}$