The net capacitance $C_{e q}=\frac{C \cdot k C}{C+k C}=\frac{C k}{1+k}$

and $Q_{e q}=C_{e q} E$

When dielectric is reemoved from capacitor $2,$ its capacitance becomes $C$.

now net capacitance $C_{e q}^{\prime}=\frac{C . C}{C+C}=\frac{C}{2}$

and $Q_{e q}^{\prime}=C_{e q}^{\prime} E=\frac{C E}{2}$

When the capacitors are connected in series, they each have the same charge as the net capacitance. Thus, $Q_{e q}=Q_{1}=Q_{2}$ and $Q_{e q}^{\prime}=Q_{1}^{\prime}=Q_{2}^{\prime}$

$\therefore \frac{Q_{1}^{\prime}}{Q_{1}}=\frac{Q_{2}^{\prime}}{Q_{2}}=\frac{2 k}{k+1}$