Answer
The energy density of the electric field increases.
Work Step by Step
Since the capacitor is connected to a battery, the potential difference $V$ across the capacitor remains the same.
We can write a general equation for the electric field strength:
$E = \frac{V}{d}$
Note that $d$ is the plate separation.
Since $d$ decreases while $V$ remains the same, the magnitude of the electric field $E$ between the plates increases.
We can write a general expression for the energy density:
$u = \frac{1}{2}\epsilon_0~E^2$
Since the magnitude of $E$ increases, then the energy density of the electric field increases.