Initially, $\quad C_{ eq }=\frac{C}{2}, V=E$

So, charge that is delivered by cell is

$Q_1=C_{ eq } E=\frac{C E}{2}$

In series charge remains same for both capacitors.

When one of capacitors is filled with a dielectric then,

$C_{ eq }=\frac{k C^2}{(C+k C)}=\left(\frac{k}{1+k}\right) C$

As battery remains connected, $V=E$. So, charge of combination

$Q_2=C_{ eq } V=\left(\frac{k}{1+k}\right) C E$

So, extra charge given by cell after insertion of dielectric is

$\Delta Q=Q_2-Q_1=\left(\frac{k}{1+k}-\frac{1}{2}\right) C E$

$=\frac{k-1}{2(k+1)} \cdot C E$