1. Initial charge on the first capacitor

The first capacitor has $C_1 = 100\,\text{pF}, \qquad V_1 = 60\,\text{V}.$

So its initial charge is $Q_{\text{initial}} = C_1 V_1 = 100\times 60 = 6000\,\text{pF·V}.$ Since $1\,\text{pF·V} = 1\,\text{pC}$, this is $Q_{\text{initial}} = 6000\,\text{pC}.$

2. After connecting the second capacitor

The battery is removed, so total charge is conserved.

Let the second capacitor be $C_2$. It is initially uncharged. After connection in parallel, both capacitors have the same final voltage.

Given final voltage across the second capacitor is $20\,\text{V}$, hence final common voltage is $V_f = 20\,\text{V}.$

3. Apply conservation of charge

Final total charge stored on both capacitors is $Q_{\text{final}} = (C_1 + C_2)V_f.$

Since charge is conserved, $Q_{\text{initial}} = Q_{\text{final}}.$ Thus, $100\times 60 = (100 + C_2)\times 20.$

4. Solve for $C_2$

$6000 = 20(100 + C_2)$ $6000 = 2000 + 20C_2$ $4000 = 20C_2$ $C_2 = 200\,\text{pF}.$

5. Match with options

The capacitance of the second capacitor is $\boxed{200\,\text{pF}}.$

So the correct option is D.