A $2\mu \text{F}$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is 

$q_i=C_{i}V=2V=q$

This charge will remains constant after switch $S$ is shifted from position $1$ to position $2$. 

$U_f=\frac{1}{2}\frac{q^2}{C_f}=\frac{q^2}{2\times 10}=\frac{q^2}{20}$

Energy dissipated $=U_i-U_f=\frac{q^2}{5}$

The energy dissipated is $80\%$ of the initial stored energy as $\%$ decrease in energy 

$$\begin{gathered} =\frac{U_i-U_f}{U_i}\times 100\% \\ =\frac{q^2/5}{q^2/4}\times 100\% \\ =\frac{4}{5}\times 100\%=80\% \end{gathered}$$