Answer
$3.0\times 10^{14}~~$ electrons travel through point $c$
Work Step by Step
We can find the equivalent capacitance of $C_1$ a,d $C_2$ which are in parallel:
$C_{12} = C_1 + C_2 = 12~\mu F$
We can find the equivalent capacitance of the three capacitors:
$\frac{1}{C_{eq}} = \frac{1}{C_{12}}+\frac{1}{C_3}$
$\frac{1}{C_{eq}} = \frac{1}{12~\mu F}+\frac{1}{12~\mu F}$
$C_{eq} = 6.0~\mu F$
We can find the charge $q$:
$q = C_{eq}~V = (6.0~\mu F)(12~V) = 72~\mu C$
Note that this charge moves through point $a$
We can find the number of electrons:
$\frac{72\times 10^{-6}~C}{1.6\times 10^{-19}~C} = 4.5\times 10^{14}$
$4.5\times 10^{14}~~$ electrons travel through point $a$
Therefore, the total number of electrons which move through point $b$ and point $c$ is also $~~4.5\times 10^{14}$
Note that the potential difference across $C_1$ and $C_2$ must be equal since they are in parallel.
We can find the ratio $\frac{q_1}{q_2}$:
$V_1 = V_2$
$\frac{q_1}{C_1} = \frac{q_2}{C_2}$
$\frac{q_1}{q_2} = \frac{C_1}{C_2}$
$\frac{q_1}{q_2} = \frac{4.0~\mu F}{8.0~\mu F}$
$\frac{q_1}{q_2} = \frac{1}{2}$
We can find the number of electrons that travel through point $c$:
$\frac{2}{3}~(4.5\times 10^{14}) = 3.0\times 10^{14}$
$3.0\times 10^{14}~~$ electrons travel through point $c$
