Why neglect friction when computing torque? There is a block of wood on a plank, with friction, and the block is moving down with uniform speed. They are asking what is the torque acting through the center of the block.  

My doubt is that since the block is moving with uniform speed that means that downward force is exactly balanced by force upward due to friction so the net torque about the center should be zero but the answer they have given takes into account the downward force $mg \sin \theta$ to calculate the torque, friction is totally neglected. 
And why is the net torque not zero?
 A: Your understanding that a change in rotational velocity requires a net torque (calculated around any convenient axis) is correct.  
The only forces acting are gravity, the normal force from the ramp, and the force of friction between the ramp and the block.  Choosing an axis across the ramp through the center of mass of the block makes life simpler;  the force of gravity (all of it) acts through this point, and thus exerts no torque.
Next, consider the friction force, that is given by $mg \sin \theta$.  This acts up the ramp, at the point of contact of the ramp and block, and thus exerts a torque around the center of mass depending on the dimensions of the block. (Tall blocks are more tippy than squat blocks)  In this case, the torque is a clockwise one.  
In order for there to be no rotation, there must be a counter-clockwise torque about the CofM.  This means that the diagram is wrong;  the normal force no longer is distributed uniformly along the base, and cannot be ignored as acting through the CofM.  In fact, the normal force shifts towards the front of the block as the ramp slope increases, and when it reaches the front corner, the rear corner lifts and the block tumbles...
A: You have to specify about what you are taking torque. Let's talk about centre of mass. 
$mg\cos\theta$ passes through it. Hence, it can produce no torque. Friction acts on lower face. It can produce a torque. But, there must be a torque to balance it!
What if the block would have wanted to rotate. It would get lifted off its edge. Then, Normal Force would act at that edge only. And hence, it would produce a torque.
What you really, get is some middle case of the extreme case I talked about it. Normal Force would be shifted away from centre of mass to produce counter torque against friction.
In the solution they have taken torque about a axis on lower face. Hence, torque due to friction is zero about that but of $mg\sin\theta$ isn't.
