A block of mass m rests on a smooth inclined plane making an angle...

The forces acting on the block are its weight $mg$ acting vertically downwards and the normal reaction $N$ acting perpendicular to the inclined plane.

Resolving the weight $mg$ into two components: $mg \sin\theta$ parallel to the inclined plane and $mg \cos\theta$ perpendicular to the inclined plane.

Since there is no motion perpendicular to the inclined plane, the forces in this direction are balanced.

Therefore, the normal reaction $N$ exerted by the plane on the block is equal to the perpendicular component of the weight.

$N = mg \cos\theta$

Since the plane is smooth, there is no frictional force. Thus, the total force exerted by the plane on the block is $mg \cos\theta$.