That's actually not the answer he is looking for. This question is referring to a case of Newton's third law of motion where he is trying to determine the action/reaction forces exterted between the boxes. This is actually one of those questions that I dread for when I was taking first year physics, but I'll show you how I would go about solving it.

The answer you're looking for is 100N

First of all, you really need to draw out the problem as free-body diagram for each individual boxes. I am going to neglect the normal and the gravitational forces since they cancel each other out and do not contribute to the solution.

I'll set out the variables

F: force applied

F2on1: force of box2 on box1

F1on2: force of box1 on box2

m1: mass of box 1

m2: mass of box 2

a: acceleration

For box 1, since it is the box that is being pushed, you have the force applied at 300N to the direction of motion. You also have an opposing force coming from box 2. This force is F of box 2 on box 1, F2on1.

For box 2, it is the box that is being pushed by box 1. This box is not being directly pushed, but rather experience the pushing force indirectly through box 1. This force is F of box 1 on box 2, F1on2.

Now you can layout the formula through Newton's second law.

__For box 1:__

F - F2on1 = m1*a

__For box 2:__

F1on2 = m2*a

You can calculate a simply by assuming the both boxes are one single box. Then you have F=(m1+m2)*a

a turns out to be 2 m/s

Plug a into the equation and you get 100N for both F1on2 and F2on1. This makes sense because they have to equal to each other, otherwise one box will begin to accelerate faster than another.