Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First, we need to re-sketch this circuit and reduce it as we see in the figures below.
We need to find the equivalent capacitance of the whole circuit, so we can find the total charge.
The easiest way to find the equivalent capacitance of this circuit is to solve it as we did in the figure below.
We have 3 in-series combinations and each combination has an equivalent capacitance of $\frac{1}{2}C$, as we see in the second circuit below.
Recall that the equivalent of in-series combinations is given by
$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\cdot\cdot\cdot$$
Now we have 3 in-parallel capacitors and the combination of these 3 is $\frac{3}{2}C$, as we see in the third circuit below.
Recall that the equivalent of in-parallel combinations is given by
$$\dfrac{1}{C_{eq}}=C_1+C_2+\cdot\cdot\cdot$$
$$\boxed{(C_{eq})_{tot}=\frac{3}{2}C}$$
$$\color{blue}{\bf [b]}$$
As we see in the first circuit below, the two points a and b are in parallel.
And both are in the middle between two identical capacitors.
So the potential difference between them is zero.
$$V_a-V_b=\color{red}{\bf 0}\;\rm V$$
