Correct Answer is Option (B)

Coin $A$ is moving in simple harmonic motion in vertical direction. Now we are assuming coin will leave contact with the platform when platform is at a distance of $x$ from the mean position which is also called amplitude.
At distance $x$ the force acting on the coin is

$mg-N=m{\omega }^{2}x$

For coin to leave contact $N=0$

$\Rightarrow mg=m{\omega }^{2}x\Rightarrow x=\frac{g}{{\omega }^2}$

$\therefore$ Option (B) is correct.