1. Use the idea of equivalent air gap

For a parallel plate capacitor, if a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between plates separated by distance $d'$, then the effective separation becomes

$d_{\text{eff}} = (d' - t) + \frac{t}{K}.$

So the capacitance is

$C = \frac{\varepsilon_0 A}{(d' - t) + t/K}.$

2. Condition for same potential difference in the same capacitor combination

The combination is maintained at the same potential difference, and the plate separation is adjusted so that the electrical effect remains unchanged. Hence the capacitance must remain the same as before insertion.

Let the original plate separation be $d$.

Initially,

$C_0 = \frac{\varepsilon_0 A}{d}.$

After inserting slab of thickness $t=3\text{ mm}$ and increasing the plate separation by $2.4\text{ mm}$,

$d' = d + 2.4\text{ mm}.$

For unchanged capacitance,

$d = (d' - t) + \frac{t}{K}.$

3. Substitute values

$d = \bigl(d + 2.4 - 3\bigr) + \frac{3}{K}.$

$d = d - 0.6 + \frac{3}{K}.$

Cancel $d$ from both sides:

$0 = -0.6 + \frac{3}{K}.$

$\frac{3}{K} = 0.6$

$K = \frac{3}{0.6} = 5.$

4. Check options

• A: $3$ ❌
• B: $4$ ❌
• C: $5$ ✅
• D: $6$ ❌

Therefore, the dielectric constant is

$\boxed{5}.$