c
$\mathrm{q}_{0}=\mathrm{C}_{0} \mathrm{V}$
$q=C V$ or $q=K C_{0} V$
$\therefore \mathrm{K}=\frac{\mathrm{q}}{\mathrm{q}_{0}}=\frac{120+60}{60}=3$
Because, the capacitor plates were initially charged to $60\, \mu C$ and on inserting the dielectric an additional charge of $120\, \mu \mathrm{C}$ also flows through the battery.