At $X=-\frac{A}{2}$

$-\frac{A}{2}=A \sin \left(\omega t_{1}+\frac{\pi}{6}\right)$

$\omega t_{1}+\frac{\pi}{6}=\sin ^{-1}(-0.5)$

$\omega t_{1}+\frac{\pi}{6}=-\frac{\pi}{6}$

$t_{1}=-\frac{\pi}{3 \omega} \ldots .(1)$

At $X=+\frac{A}{2}$

$\frac{A}{2}=A \sin \left(\omega t_{2}+\frac{\pi}{6}\right)$

$\omega t_{2}+\frac{\pi}{6}=\sin ^{-1}(0.5)$

$\omega t_{2}+\frac{\pi}{6}=\frac{\pi}{6}$

$\omega t_{2}=0$

$t_{2}=0 \ldots \ldots .(2)$

The time taken by the particle,

$T=t_{2}-t_{1}$

$T=0-\left(-\frac{\pi}{3 \omega}\right)$

$T=\frac{\pi}{3 \omega}$

The correct option is $A.$