Answer
The increase in total charge stored on the capacitors is $~~20~\mu C$
Work Step by Step
We can find the equivalent capacitance:
$\frac{1}{C_{eq}} = \frac{1}{6.0~\mu F}+\frac{1}{6.0~\mu F}$
$C_{eq} = 3.0~\mu F$
We can find the charge stored in $C_{eq}$:
$q = (3.0~\mu F)(10~V) = 30~\mu C$
Note that this charge is stored on each positive plate, so the total charge is $60~\mu C$
When one of the capacitors is squeezed so the plate separation is halved, the capacitance is doubled to $12~\mu F$
We can find the equivalent capacitance:
$\frac{1}{C_{eq}} = \frac{1}{12~\mu F}+\frac{1}{6.0~\mu F}$
$C_{eq} = 4.0~\mu F$
We can find the charge stored in $C_{eq}$:
$q = (4.0~\mu F)(10~V) = 40~\mu C$
Note that this charge is stored on each positive plate, so the total charge is $80~\mu C$
We can find the increase in total charge stored on the capacitors:
$q = 80~\mu C-60~\mu C = 20~\mu C$
The increase in total charge stored on the capacitors is $~~20~\mu C$
