Answer
$4.5\times 10^{14}~~$ electrons travel through point $d$
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 $d$
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 $d$
