$\text { Initial energy }=\frac{1}{2} CV ^2$

$=\frac{1}{2} \times 600 \times 10^{-12} \times(200)^2=12\,\mu J$

When connected to another uncharged capacitor

Charge will be equally distributed on identical capacitor

$Q ^{\prime}=\frac{ Q }{2}=6 \times 10^{-8}$

Final energy $=2 \times \frac{ Q ^{\prime 2}}{2 C }=\frac{ Q ^{\prime 2}}{ C }$

$\frac{\left(6 \times 10^{-8}\right)^2}{600 \times 10^{-12}}=6\,\mu J$

Energy lost = Initial energy - Final energy

$=(12-6)\,\mu\, J =6\, \mu\, J$