Two cubic blocks are in contact, resting on a frictionless horizontal surface. The block on the left has a mass of mL = 6.70 kg, and the block on the right has a mass of mR = 18.4 kg. A force of magnitude 112 N is applied to the left face of the left block, toward the right but at an upward angle of 30.0° with respect to the horizontal. It causes the left block to push on the right block. What are (a) the direction and (b) the magnitude of the force that the right block applies to the left block?

Answer 1

x-component of the force F(x) =
=F•cos30º97 N.
F(x) = (ML +MR) •a,
a =F(x)/ (ML +MR) =
=97/(6.7+18.4) = 3.86 m/s².
F(R→L) =MR•a = 18.7•3.86 = 71.1 N.

Answer 2

The first thing I would ask is about the interface between the blocks, frictionless or not. If the interface is frictionless, the left block slides to the right and UP, and the right moves to the right. IN THAT CASE, consider this:

left horizontal force: 112cos30=figure it
acceleartion to the left of both blocks
horizontal force=(sumofmasses)a

force right block pushes back on the left: MR*a

Now, if the interface between the blocks is friction, it matters how much friction. If they were "glued", the upward force on the left (112*sin30) would then move both up, and frankly, start the system spinning about the center of mass. Probably not within your physics capability yet to figure.
If the friction is less, so that the right block does not move up, then that friction is acting on the left block downward. It has to be added to the horizontal force the right ispushing back on.

SO I don't understand what is assumed here.