A 3.0 µF capacitor and a 6.0 µF capacitor are first joined in parallel across a 60 V battery…

Step 1. In parallel. Each has $60\ \text{V}$. $U_1 = \tfrac{1}{2} \times 3.0 \times 10^{-6}\ \text{F} \times (60\ \text{V})^2 = 5.4\ \text{mJ}$ and $U_2 = \tfrac{1}{2} \times 6.0 \times 10^{-6}\ \text{F} \times (60\ \text{V})^2 = 10.8\ \text{mJ}$.

Step 2. In series, the charge. $C = \dfrac{3.0\ \mu\text{F} \times 6.0\ \mu\text{F}}{9.0\ \mu\text{F}} = 2.0\ \mu\text{F}$, so $Q = 2.0\ \mu\text{F} \times 60\ \text{V} = 120\ \mu\text{C}$ on each.

Step 3. In series, the energies. $U_1 = \dfrac{(120 \times 10^{-6}\ \text{C})^2}{2 \times 3.0 \times 10^{-6}\ \text{F}} = 2.4\ \text{mJ}$ and $U_2 = \dfrac{(120 \times 10^{-6}\ \text{C})^2}{2 \times 6.0 \times 10^{-6}\ \text{F}} = 1.2\ \text{mJ}$.

What to learn. In parallel the bigger capacitor stores more. In series the smaller one does.