Answer: $F\large\frac{m_2+m_3}{m_1+m_2+m_3}$

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

Here, R' is the contact force on $m_2$ and is the reaction force between $m_1$ and $m_2$

Similarly, R is the contact force on $m_3$ and is the reaction force between $m_2$ and $m_3$

For block-1, $F - R' = m_1a$ and for block-2, $R'-R = m_2a$ and for block-3, $R = m_3a$

Adding the three equations, $F - R' + R' - R + R = F = (m_1+ m_2+m_3)a$

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

Therefore we can solve for $R' = m_2 a+R = (m_2+m_3)a=F\large\frac{m_2+m_3}{m_1+m_2+m_3}$