For capacitors in series, the charge $Q$ on each capacitor is identical:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4 + 2 + 1}{12} = \frac{7}{12}\,\mu\text{F}^{-1}$$ $$C_{\text{eq}} = \frac{12}{7}\,\mu\text{F}$$

The total charge supplied by the $7\,\text{V}$ battery is:

$$Q = C_{\text{eq}} V = \left(\frac{12}{7}\,\mu\text{F}\right) \times 7\,\text{V} = 12\,\mu\text{C}$$

The potential difference across the $6\,\mu\text{F}$ capacitor ($C_2$) is:

$$V_2 = \frac{Q}{C_2} = \frac{12\,\mu\text{C}}{6\,\mu\text{F}} = 2\,\text{V}$$

Copper is a metal (conductor) with an infinite dielectric constant ($K \to \infty$).

The capacitance of a parallel plate capacitor with a slab of thickness $b$ and dielectric constant $K$ is:

$$C = \frac{\varepsilon_0 A}{d - b + \frac{b}{K}}$$

For a metallic conductor like copper, $K \to \infty$, which gives $\frac{b}{K} \to 0$:

$$C = \frac{\varepsilon_0 A}{d - b}$$

The effective separation between the capacitor plates is reduced from $d$ to $(d - b)$. Hence, option 2 is the correct answer.