There is no loop (closed circuit) in this problem, so you cannot use Faraday's Law. You must use the fact that the emf across a wire of length L moving at velocity v perpendicular to a magnetic field B is BLv.

The difficulty is that different segments of the conducting wire move at different speeds. So you must integrate from the pivot to the bob, adding the contribution from successive segments.

For all points on the wire, the maximum speed occurs when the wire passes through the vertical.

The amplitude of swing is $\theta$ so the maximum speed of the bob is given by

$v_0^2=2gh=2gL(1-\cos\theta)=4gL\sin^2\frac{\theta}{2}$

$v_0=2\sqrt{gL}\sin\frac{\theta}{2}$.

The emf induced in a wire of length $L$ moving with speed $v$ perpendicular to a field $B$ is $BLv$. Speed $v$ in the pendulum wire varies uniformly from 0 to $v_0$ so the average speed is $\frac12 v_0$. Because $V \propto v$ and $v$ varies uniformly we can use average speed instead of integrating.

The total emf induced in the pendulum wire is

$V=\frac12 BLv_0=BL\sqrt{gL}\sin\frac{\theta}{2}$.

**Note:** All of the options have different dimensions. So a quick way of getting the answer is to check which option has the dimensions of voltage - assuming that this option is correct.