1. Initial capacitance

For a parallel plate capacitor of plate separation $d=5\text{ mm}$ and plate area $A$, the initial capacitance is

2. Battery remains connected

Since the battery remains connected, the potential difference $V$ across the capacitor stays constant.

So charge is proportional to capacitance:

Given that the capacitor draws $25\%$ more charge after inserting the dielectric,

Hence,

3. Capacitance after inserting dielectric slab

A dielectric slab of thickness $t=2\text{ mm}$ and dielectric constant $K$ is inserted between plates separated by $d=5\text{ mm}$.

This is equivalent to an effective air gap of

So the new capacitance is

Substitute $d=5$ mm and $t=2$ mm:

4. Use the given condition

We know

That is,

Cancel $\varepsilon_0 A$:

So,

5. Final answer

The dielectric constant of the sheet is

6. Comparison with stored answer

Stored correct answer = $2$

Our derived answer = $2$

So the answer agrees with the stored correct answer.