Step 1: Understand the setup.
We have a parallel plate capacitor with air between the plates, charged to $100\ \text{V}$. A dielectric slab of thickness $2\ \text{mm}$ is slid in. To keep the same voltage (same capacitance for the same charge), the plates are moved apart by $1.6\ \text{mm}$. We must find the dielectric constant $k$ of the slab.
Step 2: Why moving the plates helps.
Putting in a dielectric raises the capacitance. To bring the capacitance back to its old value, we increase the gap between the plates. A bigger gap lowers capacitance. So the two effects balance out.
Step 3: Capacitance with air.
With plain air and gap $d$:
\[ C = \frac{\epsilon_0 A}{d} \]
Step 4: Capacitance with the slab and bigger gap.
When a slab of thickness $t$ and dielectric constant $k$ sits inside a gap, the slab acts like a smaller air gap of size $t/k$. The new total gap is $d + x$ where $x = 1.6\ \text{mm}$ is the extra distance. So the effective gap is:
\[ (d + x) - t + \frac{t}{k} \]
Step 5: Match the two capacitances.
Since capacitance is the same, the effective gaps must be equal:
\[ d = (d + x) - t + \frac{t}{k} \]
Cancel $d$ from both sides:
\[ 0 = x - t + \frac{t}{k} \]
Step 6: Solve for $k$.
Rearrange to $t - x = \dfrac{t}{k}$. Put $t = 2\ \text{mm}$ and $x = 1.6\ \text{mm}$:
\[ 2 - 1.6 = \frac{2}{k} \quad\Rightarrow\quad 0.4 = \frac{2}{k} \]
\[ k = \frac{2}{0.4} = 5 \]
This matches option (2).
\[ \boxed{k = 5} \]