Answer
$V_A-V_B = 5.25~V$
Work Step by Step
We can find the potential difference across $C_1$:
$V_1 = \frac{30~\mu C}{10~\mu F} = 3.0~V$
Since $C_2$ is in parallel with $C_1$. the potential difference across $C_2$ is also $3.0~V$
We can find the charge on $C_2$:
$q = C_2~V_2 = (20~\mu F)(3.0~V) = 60~\mu C$
The total charge on $C_1$ and $C_2$ is $90~\mu C$. This must also be the total charge on $C_3$ and $C_4$
Since $C_3=C_4$, by symmetry, they must each have half of the total charge.
We can find the potential difference across $C_3$:
$V_3 = \frac{q_3}{C_3} = \frac{45~\mu C}{20~\mu F} = 2.25~V$
We can find the magnitude of the potential difference $V_A - V_B$:
$V_A-V_B = 3.0~V+2.25~V = 5.25~V$
