A cubical block of mass m and edge length a is sliding down a rough...

Since the block is sliding down the inclined plane at a uniform speed, its net acceleration and net angular acceleration are zero.

The forces acting on the block are its weight $mg$, the normal force $N$, and the frictional force $f$.

For translational equilibrium along the inclined plane:

$f = mg \sin\theta$

The block is in rotational equilibrium, so the net torque about its centre of mass must be zero. The weight $mg$ passes through the centre of mass, so its torque is zero.

The frictional force $f$ acts along the bottom surface of the block, which is at a perpendicular distance of $\dfrac{a}{2}$ from the centre of mass. The torque due to friction about the centre is:

$\tau_f = f \times \dfrac{a}{2} = \dfrac{1}{2}mga \sin\theta$

To maintain rotational equilibrium, the torque due to the normal force $\tau_N$ about the centre must be equal and opposite to the torque due to friction.

$\tau_N = \tau_f = \dfrac{1}{2}mga \sin\theta$