Answer
$q = 1.06\times 10^{-9}~C$
Work Step by Step
We can find the capacitance of the slab with dielectric constant $\kappa = 3.00$:
$C = \frac{\kappa~\epsilon_0~A}{d}$
$C = \frac{(3.00)(8.854\times 10^{-12}~F/m)(2.00\times 10^{-2}~m^2)}{2.00\times 10^{-3}~m}$
$C = 265.62\times 10^{-12}~F$
We can find the capacitance of the slab with dielectric constant $\kappa = 4.00$:
$C = \frac{\kappa~\epsilon_0~A}{d}$
$C = \frac{(4.00)(8.854\times 10^{-12}~F/m)(2.00\times 10^{-2}~m^2)}{2.00\times 10^{-3}~m}$
$C = 354.16\times 10^{-12}~F$
Note that the two slabs are in series.
We can find the equivalent capacitance:
$\frac{1}{C_{eq}} = \frac{1}{265.62\times 10^{-12}~F}+\frac{1}{354.16\times 10^{-12}~F}$
$C_{eq} = 151.8\times 10^{-12}~F$
We can find the charge stored on the capacitor:
$q = C_{eq}~V$
$q = (151.8\times 10^{-12}~F)(7.00~V)$
$q = 1.06\times 10^{-9}~C$
