Answer
$10~\mu C~~$ is transferred from the battery.
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 $C_{eq}$:
$q = (3.0~\mu F)(10~V) = 30~\mu C$
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 $C_{eq}$:
$q = (4.0~\mu F)(10~V) = 40~\mu C$
We can find the additional charge that is transferred from the battery:
$q = 40~\mu C-30~\mu C = 10~\mu C$
$10~\mu C~~$ is transferred from the battery.
