Work done $W = MB(1 - \cos\theta) = MB(1 - \cos 60^{\circ}) = MB/2$. Torque $\tau = MB\sin\theta = MB\sin 60^{\circ} = MB(\sqrt{3}/2)$. From the work equation, $MB = 2W$. Substituting this into the torque equation gives $\tau = (2W)(\sqrt{3}/2) = \sqrt{3}W$.