Answer
$\Sigma \tau = mg \mathcal{l}_2-mg \mathcal{l}_1$ $clockwise$.
Work Step by Step
There is a torque due to the weight on the left that tends to accelerate the rod counterclockwise and a clockwise torque due to the weight on the right. Let’s call clockwise the positive direction.
$$\Sigma \tau = mg \mathcal{l}_2-mg \mathcal{l}_1$$
The net torque is clockwise.