A capacitor C1=2 muF is charged to 100 V and a capacitor C2=4 muF is charged to 60 V. They are then joined with the positive plate of one connected…

Before joining, $q_1=2\times100=200\ \mu\mathrm{C}$ and $q_2=4\times60=240\ \mu\mathrm{C}$. Joining unlike plates makes the charges partly cancel, so the net charge on the connected pair of plates is $q=240-200=40\ \mu\mathrm{C}$. After charge flow stops the capacitors are in parallel with total capacitance $6\ \mu\mathrm{F}$ and potential difference $40/6=6.67\ \mathrm{V}$. With the pair isolated the net charge stays $40\ \mu\mathrm{C}$. Filling $C_1$ with the dielectric makes $C_1'=6\ \mu\mathrm{F}$, so the parallel capacitance becomes $10\ \mu\mathrm{F}$ and $V=4\ \mathrm{V}$. The energy is $U=\dfrac{q^2}{2C}=\dfrac{(40)^2}{2\times10}=80\ \mu\mathrm{J}$. The wrong choices are: $133\ \mu\mathrm{J}$ (forgetting the dielectric), $222\ \mu\mathrm{J}$ (keeping the voltage instead of the charge fixed) and $9680\ \mu\mathrm{J}$ (adding the charges as if the like plates had been joined).