$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

splitting $\frac{2 \pi}{3}$ as $\frac{(3 \pi-\pi)}{3}$ or $\pi-\frac{\pi}{3}$

and substitute the same in given equation we get,
$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)=\frac{\pi}{3}$

Therefore, $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ is $\frac{\pi}{3}$