Answer: $\large\frac{Fm_2}{m_1+m_2}$

Lets draw the Free Body Diagrams for the two masses in contact and the resultant forces:

Here R is the normal reaction force between the blocks.

For block-1, $F - R = m_1a$ and for block-2, $R = m_2a$

Adding the two equations, $F - R + R = F = (m_1+ m_2)a$

$\Rightarrow a = \large\frac{F}{m_1+m_2}$

Therefore we can solve for $R = m_2 a = \large\frac{Fm_2}{m_1+m_2}$