If the charge on the capacitor is increased by 2C, the energy stored in it increases by $21 \%$. The original charge on the capacitor is

The original charge $Q$ and capacitance $C$ yield stored energy $U = \frac{Q^2}{2C}$.

After adding $2\text{ C}$, the charge becomes $Q + 2$ and energy becomes $U' = \frac{(Q + 2)^2}{2C}$.

Given a $21\%$ energy increase, $U' = 1.21U$, so

\[\frac{(Q + 2)^2}{2C} = 1.21 \cdot \frac{Q^2}{2C}.\]

Canceling $\frac{1}{2C}$ from both sides, $(Q + 2)^2 = 1.21Q^2$.

Solving this quadratic equation gives $Q + 2 = \pm 1.1Q$.

The positive case yields $Q = \frac{2}{0.1} = 20\text{ C}$, while the negative case gives $Q = -\frac{2}{2.1}\text{ C}$, which is discarded.

Final answer: $20\text{ C}$