Answer: $\large\frac{5}{3}$

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

Here, R' is the contact force on m2m_2 and is the reaction force between m1m_1 and m2m_2

Similarly, R is the contact force on m3m_3 and is the reaction force between m2m_2 and m3m_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 $F_3=R= m_3a = F\large\frac{m_3}{m_1+m_2+m_3}$

Therefore we can solve for $F_2=R'= m_2a+m_3a = F\large\frac{m_2+m_3}{m_1+m_2+m_3}$

Given that the masses are in the ratio $1:2:3$, $F_3 = F \large\frac{3m}{m+2m+3m}$$ = \large\frac{3F}{6}$

Given that the masses are in the ratio $1:2:3$, $F_2 = F \large\frac{2m+3m}{m+2m+3m}$$ = \large\frac{5F}{6}$

$\Rightarrow$ Ratio of $F_2:F_3$ is $\large\frac{F_2}{F_3}$$ = \Large\frac{\frac{5F}{6}}{\frac{3F}{6}}$$ = \large\frac{5}{3}$