1. Given data

• Plate area: $A = 40\,\text{cm}^2 = 40\times 10^{-4}\,\text{m}^2 = 4\times 10^{-3}\,\text{m}^2$
• Total plate separation: $d = 2\,\text{mm}$
• Dielectric slab thickness: $t = 1\,\text{mm}$
• Remaining air gap: $d-t = 1\,\text{mm}$
• Dielectric constant: $K=5$

2. Concept used

Since the dielectric fills only part of the separation, the system acts like two capacitors in series:

• one with air thickness $1\,\text{mm}$
• one with dielectric thickness $1\,\text{mm}$ and dielectric constant $5$

For layered media along the separation, the equivalent capacitance is:

$C = \frac{\varepsilon_0 A}{d_\text{air} + \dfrac{d_\text{dielectric}}{K}}$

3. Substitute values

Here,

$d_\text{air} = 1\,\text{mm} = 10^{-3}\,\text{m}$ $d_\text{dielectric} = 1\,\text{mm} = 10^{-3}\,\text{m}$

So,

$C = \frac{\varepsilon_0 (4\times 10^{-3})}{10^{-3} + \dfrac{10^{-3}}{5}}$

$= \frac{\varepsilon_0 (4\times 10^{-3})}{10^{-3}\left(1+\frac15\right)}$

$= \frac{\varepsilon_0 (4\times 10^{-3})}{10^{-3}\cdot \frac65}$

$= \varepsilon_0 \cdot 4 \cdot \frac{5}{6}$

$= \frac{20}{6}\varepsilon_0$

$= \frac{10}{3}\varepsilon_0\,\text{F}$

4. Option check

• A: $10\varepsilon_0$ ❌
• B: $24\varepsilon_0$ ❌
• C: $\dfrac{3}{10}\varepsilon_0$ ❌
• D: $\dfrac{10}{3}\varepsilon_0$ ✅

Therefore, the correct answer is:

$\boxed{\frac{10}{3}\varepsilon_0\,\text{F}}$