all the surfaces are frictionless. find the force on the sphere by inclined surface.

Draw the free Body diagram first.
mg, gravitational force / weight of the body acting downwards,
n, normal reaction force due to the inclined part of the mother block with it's line of action passing through the centre of the sphere inclined at an angle 30 deg. with the upward vertical axis,
N, normal reaction force due to the vertical part of the mother block with it's line of action passing through the centre of sphere,
ma, pseudo force acting on the sphere as observed from the mother block frame of reference.

Now, The question requires us to find the net reaction force exerted by the inclined part of the mother block on to the sphere.

We know that, Net reaction force exerted by a surface (R) =√(RN​2+Rf2)
where, RN = normal force exerted by the surface
and, Rf = frictional force exerted by the surface.
 

Now, for the inclined part, it is given that Rf = zero and,
we have assigned RN = n
Now, as the sphere is in vertical and horizontal equilibrium (with respect to the reference frame where I am in),
the vertical component of all forces must algebraically add up to zero,
i.e., n*cos(30o) = mg
or, n = 2mg/√(3)