A parallel-plate capacitor possessing a plate area of 25.0 cm^2 and...

Initial capacitance $C_1 = \dfrac{\varepsilon_0 A}{d_1}$

Final capacitance $C_2 = \dfrac{\varepsilon_0 A}{d_2}$

Since the battery remains connected, the potential difference $V$ across the capacitor is constant.

Initial charge $Q_1 = C_1 V = \dfrac{\varepsilon_0 A V}{d_1}$

Final charge $Q_2 = C_2 V = \dfrac{\varepsilon_0 A V}{d_2}$

Extra charge supplied \Delta Q = Q_2 - Q_1 = \varepsilon_0 A V \left(\dfrac{1}{d_2} - \dfrac{1}{d_1} \right)

Substituting the given values:
$A = 25.0 \times 10^{-4} \text{ m}^2$
$d_1 = 2.00 \times 10^{-3} \text{ m}$
$d_2 = 1.00 \times 10^{-3} \text{ m}$
$V = 12.0 \text{ V}$
$\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

$\Delta Q = 8.85 \times 10^{-12} \times 25.0 \times 10^{-4} \times 12.0 \times \left( \dfrac{1}{1.00 \times 10^{-3}} - \dfrac{1}{2.00 \times 10^{-3}} \right)$

$\Delta Q = 265.5 \times 10^{-15} \times (1000 - 500)$

$\Delta Q = 265.5 \times 10^{-15} \times 500 = 1.3275 \times 10^{-10} \text{ C}$

Rounding to three significant figures, we get $\Delta Q \approx 1.33 \times 10^{-10} \text{ C}$