Answer
$6.7\times10^{-7}C $.
Work Step by Step
Use the capacitance equation, Q = CV.
The initial charge on the capacitor is $Q_i = C_iV$. After the mica is inserted, the capacitance increases by a factor of K, and the voltage is unchanged because the battery is still attached to the capacitor. The final charge on the capacitor can be found by $Q_f=C_fV$.
$$\Delta Q = Q_f-Q_i= C_fV - C_iV $$
$$\Delta Q = KC_iV - C_iV=(K-1) C_iV=(7-1)(3500\times10^{-12}F)(32V)=6.7\times10^{-7}C $$