Answer
$3.7\times10^{-6}C $.
Work Step by Step
The energy stored in the capacitor is $PE=\frac{Q^2}{2C}$.
The final energy is half the initial energy. Find the final charge.
$$ \frac{Q_f^2}{2C}=\frac{1}{2}\frac{Q_i^2}{2C}$$
$$Q_f=\frac{1}{\sqrt{2}}Q_i$$
Subtract the final charge from the initial charge to find the charge that leaked off.
$$Q_{leak}=Q_i-Q_f=Q_i(1-\frac{1}{\sqrt{2}})=CV(1-\frac{1}{\sqrt{2}})$$
$$Q_{leak}= (2.1\times10^{-6}F)(6.0V)(1-\frac{1}{\sqrt{2}})=3.7\times10^{-6}C $$
