Find the equivalent capacitance between A and B

$q_1+q_2=q$…………..(1)

Using KVL, 

$-\frac{q_1}{C}-\frac{q_3}{2C}+\frac{q_2}{2C}=0$

$\Rightarrow q_2-q_3-2q_1=0$………(2)

Plates inside dotted line form an isolated system.

Hence, $q_2+q_3-q_1=0$………..(3)

Solving these equations, we get

$q_1=\frac{2}{5}q,q_2=\frac{3}{5}q\text{ and }{q}_3=-\frac{q}{5}$

$\text{ Now, let }{C}_{\mathrm{eq}}\text{ be the equivalent capacitance between }A\text{ and }B\text{ . }$

$\begin{array}{l}{V}_A-V_B=\frac{q}{C_{\text{eq}}}=\frac{q_1}{C}+\frac{q_2}{2C}\\ \Rightarrow \frac{q}{C_{\text{eq}}}=\frac{2q}{5C}+\frac{3q}{10C}=\frac{7q}{10C}\\ \Rightarrow C_{\text{eq}}=\frac{10}{7}C\end{array}$