The correct answer is:
$$
\begin{aligned}
\text { (a) : } W &=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \\
&=-M B\left(\cos 60^{\circ}-\cos 0\right)=\frac{M B}{2} \\
\therefore \quad & M B=2 W \\
& \text { Torque }=M B \sin 60^{\circ}=(2 W) \sin 60^{\circ} \\
=& \frac{2 W \times \sqrt{3}}{2}=\sqrt{3} W .
\end{aligned}
$$