The given circuit forms a Wheatstone bridge. But the bridge is not balanced. Let us suppose point `A` is connected to the positive terminal of a battery and `B` to the negative terminal of the same battery, so that a total charge q is stored in the capacitors. Just by seeing input and output symmetry, we can say that charges will be distributed as shown below.
`q_1+q_2=q` ............i
Applying second law, we have
`-q_i/C-q_3/(2C)=0`
or `q_2-q_3-2q_1=0`...............ii
Plates inside the dotted line form an isolated system. Hence,
`q_2+q_3-q_1=0` ..............iii
Solving these three equations, we have
`q_1=2/5q, q_2=3/5q` and `q_3=-q/5`
Now let `C_(eq)` be the equivalent capacitance between `A` and `B`. Then,
`V_A-V_B=q/(C_eq)=q_1/C+q_2/(2C)`
`:. q/C_(eq)=(2q)/(5C)+(3q)/(10C)=(7q)/(10C)`
`:. C_(eq)=10/7C`
