A block of mass m is placed on a smooth we

When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.

For the condition of equilibrium

$mg\,\sin \theta = ma\cos \theta $ $⇒$ $a = \frac{{g\sin \theta }}{{\cos \theta }}$

$\therefore $ Force exerted by the wedge on the block

$R = mg\cos \theta + ma\sin \theta $

$R = mg\cos \theta + m\left( {\frac{{g\sin \theta }}{{\cos \theta }}} \right)\sin \theta $

$ = \frac{{mg({{\cos }^2}\theta + {{\sin }^2}\theta )}}{{\cos \theta }}$

$R = \frac{{mg}}{{\cos \theta }}$