The initial force is $F = k \frac{(Q)(2Q)}{r^2} = \frac{2kQ^2}{r^2}$.

Step 1: C touches A. Since they are identical, they share the total charge equally.

$$q_A' = q_C' = \frac{q_A + q_C}{2} = \frac{+Q - 2Q}{2} = -\frac{Q}{2}$$

Step 2: C (with charge $q_C'$) touches B. They share the total charge equally.

$$q_B' = q_C'' = \frac{q_B + q_C'}{2} = \frac{+2Q - Q/2}{2} = \frac{3Q/2}{2} = +\frac{3Q}{4}$$

Final state: The charges are $q_A' = -Q/2$ and $q_B' = +3Q/4$. The new distance is $r' = r/2$. The new force $F'$ is:

$$F' = k \frac{|q_A' q_B'|}{(r')^2} = k \frac{|(-Q/2)(3Q/4)|}{(r/2)^2} = k \frac{3Q^2/8}{r^2/4} = k \frac{12Q^2}{8r^2} = \frac{3}{2}\frac{kQ^2}{r^2}$$

To find the ratio $F'/F$:

$$\frac{F'}{F} = \frac{\frac{3}{2}\frac{kQ^2}{r^2}}{2\frac{kQ^2}{r^2}} = \frac{3/2}{2} = \frac{3}{4}$$

Therefore, $F' = \frac{3}{4}F$.