Use the energy formula for a capacitor

For a capacitor of fixed capacitance $C$, the energy stored is $U = \frac{Q^2}{2C}$

Let the original charge be $Q$

After increasing the charge by $2\,\text{C}$, the new charge becomes $(Q+2)$

So the new energy is $U' = \frac{(Q+2)^2}{2C}$

Use the given percentage increase in energy

The energy increases by $44\%$, so $U' = 1.44U$

Therefore, $\frac{(Q+2)^2}{2C} = 1.44\cdot \frac{Q^2}{2C}$

Simplify

Cancel $\frac{1}{2C}$ from both sides: $(Q+2)^2 = 1.44Q^2$

Taking square root: $Q+2 = 1.2Q$

since charge is positive.

Solve for $Q$

$2 = 1.2Q - Q = 0.2Q$ $Q = \frac{2}{0.2} = 10$

Check with options

The original charge is $\boxed{10\,\text{C}}$

So the correct option is A.