Battery disconnected → charge $Q$ remains constant throughout.

(a) Capacitance: with distance doubled ($2d$) and dielectric $K=1.8$ filling completely:

$$C'=\frac{K\varepsilon_0A}{2d}=\frac{K}{2}\times\frac{\varepsilon_0A}{d}=\frac{1.8}{2}C=0.9C$$

Since $0.9<1$, the capacitance DECREASES (to 90% of original).

(b) Electric field: with $Q$ constant, $E'=\dfrac{Q}{K\varepsilon_0A}=\dfrac{E_0}{K}=\dfrac{E_0}{1.8}$

The field DECREASES (note: it depends only on $Q$, $A$, $K$ — not on the plate separation $d$).

(c) Energy stored: $U=\dfrac{Q^2}{2C}$. Since $Q$ is constant and $C$ decreases to $0.9C$:

$$U'=\frac{Q^2}{2(0.9C)}=\frac{U_0}{0.9}\approx1.11\,U_0$$

The energy INCREASES.