Answer
$q = 1.06\times 10^{-9}~C$
Work Step by Step
We can find the capacitance of the capacitor filled with air:
$C_2 = \frac{\epsilon_0~A}{d}$
$C_2 = \frac{(8.854\times 10^{-12}~F/m)(5.00\times 10^{-3}~m^2)}{2.00\times 10^{-3}~m}$
$C_2 = 22.135\times 10^{-12}~F$
We can find the capacitance of the capacitor with the dielectric:
$C_1 = \frac{\kappa \epsilon_0~A}{d}$
$C_1 = \frac{(3.00)(8.854\times 10^{-12}~F/m)(5.00\times 10^{-3}~m^2)}{2.00\times 10^{-3}~m}$
$C_1 = 66.405\times 10^{-12}~F$
We can find the equivalent capacitance:
$C_{eq} = C_1+C_2$
$C_{eq} = (66.405\times 10^{-12}~F)+(22.135\times 10^{-12}~F)$
$C_{eq} = 88.54\times 10^{-12}~F$
We can find the charge stored on the capacitors:
$q = C_{eq}~V$
$q = (88.54\times 10^{-12}~F)(12.0~V)$
$q = 1.06\times 10^{-9}~C$
