Answer
The correct answer is:
(c) $(\frac{mg}{2})~sin~\theta$
Work Step by Step
In general, the torque is $\tau = r\times F$, where $r$ is the smallest distance between the force vector and the axis, and $F$ is the perpendicular component of the force vector.
To hold the bar in equilibrium, the force F must produce a torque that is equal in magnitude to the torque produced by gravity.
Let $L$ be the length of the bar. Note that we can consider that the force of gravity is applied at the midpoint of the bar. We can find the magnitude of the torque produced by gravity:
$\tau_g = mg~(\frac{L}{2})~sin~\theta$
We can equate the magnitudes of the torques to find $F$:
$L\times F = mg~(\frac{L}{2})~sin~\theta$
$F = (\frac{mg}{2})~sin~\theta$
The correct answer is:
(c) $(\frac{mg}{2})~sin~\theta$
