A block of mass m is positioned on a smooth wedge having an...

The block does not slip on the wedge, which implies it moves horizontally with the same acceleration as the wedge.

The forces acting on the block are its weight $mg$ acting vertically downwards and the normal reaction $N$ exerted by the smooth wedge, acting perpendicular to the inclined surface.

Since there is no vertical acceleration of the block, the vertical component of the normal reaction balances the weight of the block.

Equating the vertical forces:

$N \cos\theta = mg$

$N = \dfrac{mg}{\cos\theta}$

The magnitude of the force exerted on the block by the wedge is the normal reaction $N$, which is $\dfrac{mg}{\cos\theta}$.