Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is (in pF):

The first capacitor has $$C_1 = 100\,$$pF and is initially charged to $$V_{1i}=60\,$$V.

Initial charge on this capacitor:
$$Q_{\text{initial}} = C_1\,V_{1i} = 100\,\text{pF}\times 60\,\text{V} = 6000\,\text{pC}$$

After disconnecting the battery, this charge cannot escape; it is conserved inside the closed system formed by the two capacitors.

An uncharged capacitor of capacitance $$C_2$$ is then connected in parallel. In a parallel connection the final voltage on both capacitors is the same. The problem states that this common voltage is $$V_f = 20\,$$V.

Let $$Q_{1f}$$ and $$Q_{2f}$$ be the final charges on $$C_1$$ and $$C_2$$ respectively.
Using $$Q = C\,V$$:
$$Q_{1f} = C_1\,V_f$$
$$Q_{2f} = C_2\,V_f$$

Because total charge is conserved (battery is removed),
$$Q_{\text{initial}} = Q_{1f} + Q_{2f}$$
$$\Rightarrow 6000 = C_1\,V_f + C_2\,V_f$$

Divide both sides by $$V_f$$ to isolate $$C_2$$:
$$\frac{6000}{V_f} = C_1 + C_2$$
$$\Rightarrow C_2 = \frac{6000}{20} - C_1$$

Substitute the known values:
$$C_2 = 300 - 100 = 200\,\text{pF}$$

Therefore, the capacitance of the second capacitor is $$200\,$$pF, which corresponds to Option B.