A cubical block of mass M and edge length a slides down a rough...

Since the cubical block moves down the inclined plane with uniform velocity, its translational and rotational accelerations are zero.

For translational equilibrium along the inclined plane, the frictional force $f$ balances the component of weight:

$f = Mg\sin\theta$

For rotational equilibrium about the centre of the block, the net torque must be zero. The forces acting on the block are weight $Mg$, normal force $N$, and friction $f$.

The weight $Mg$ acts through the centre, so its torque about the centre is zero.

The frictional force $f$ acts along the bottom surface, which is at a perpendicular distance of $\dfrac{a}{2}$ from the centre. 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 produced by the normal force $N$ about the centre must balance the torque produced by friction. Therefore, the magnitude of the torque due to the normal force is equal to the magnitude of the torque due to friction:

$\tau_{N} = \tau_{f} = \dfrac{1}{2}Mga\sin\theta$