Answer
See explanation.
Work Step by Step
The electric field is stronger for the capacitor with the plates that are closer. The potential difference is the same. The potential difference is the electric field strength multiplied by the distance between the parallel plates, so the one with smaller d has the stronger E.
The charge is greater for the capacitor with the plates that are closer. That capacitance is larger, because for a parallel-plate capacitor, $C=\frac{\epsilon_oA}{d}$. The charge is $Q=CV$, so this capacitance is larger, while both capacitors have the same potential difference. Therefore the charge is greater on the capacitor with the plates that are closer.
The energy density is greater for the capacitor with the plates that are closer, because $u=\frac{\epsilon_o}{2}E^2$ and we know that capacitor has a stronger electric field.
