Answer
$\tau = 4.6~N\cdot m$
Work Step by Step
We can write a general expression for the torque about the pivot:
$\tau = r\times F = r~F~sin~\theta$
$r$ is the displacement vector from the pivot to the point where the force is applied
$F$ is the force vector
$\theta$ is the angle between these two vectors
When the pendulum is $30^{\circ}$ from the vertical, then the angle between the $r$ vector and the force vector is $150^{\circ}$
We can find the magnitude of the gravitational torque about the pivot:
$\tau = r~F~sin~\theta$
$\tau = r~mg~sin~\theta$
$\tau = (1.25~m)(0.75~kg)(9.8~m/s^2)~sin~150^{\circ}$
$\tau = 4.6~N\cdot m$
