A cubical block of mass M and edge a slides down a rough inclined plane of inclination $\mathrm{\theta }$ with a uniform velocity. The torque of the normal force on the block about its center has a magnitude of $\frac{\mathrm{aMgsin\theta }}{\mathrm{K}}$. The value of K is ______

In this case the normal reaction will not pass through center, as shifting of normal takes place.

As the block is moving constant velocity, the total net torque acting on the body will be zero.

So about the center of block torque produced by the friction force is balanced by the torque produced by the normal reaction.

The torque by frictional force about center of mass is given by 

$\mathrm{\tau }=\frac{\mathrm{a}}{2}\times \mathrm{Mgsin\theta }=\frac{\mathrm{aMgsin\theta }}{2}$