An air capacitor of 60 pF is charged to 90 V…

A dielectric that fills the gap multiplies $C$ by $K$. With the battery connected $V$ stays and $Q$ grows by $K$. With the battery removed $Q$ stays and $V$ falls to $V_0/K$. A slab of thickness $t$ gives $C = \dfrac{\varepsilon_0 A}{d - t + t/K}$, and the field in the slab is $1/K$ of the field in the air. A metal slab gives $C = \dfrac{\varepsilon_0 A}{d - t}$.