A parallel-plate capacitor with plate area A and separation d has capacitance C0 and stays connected to a battery of emf V0. A dielectric slab of…

The slab and the air layer act as two capacitors in series, so the final capacitance is $C=\varepsilon_0A\big/\left(\tfrac d2+\tfrac{d}{2\kappa}\right)=\dfrac{\varepsilon_0A}{2d/3}=1.5\,C_0$; (A) is true. The charge is $Q=1.5C_0V_0$, so the field in the air gap is $E_a=Q/\varepsilon_0A=1.5V_0/d$, and (B) is true. The field in the dielectric is $E_a/\kappa=0.5V_0/d$, making (D) false; as a check, $E_a\frac d2+E_d\frac d2=0.75V_0+0.25V_0=V_0$. Energy balance: the battery supplies $\Delta Q\,V_0=0.5C_0V_0^2$, the stored energy rises by $\tfrac12(1.5-1)C_0V_0^2=0.25C_0V_0^2$, so the external agent's work is $0.25-0.50=-0.25\,C_0V_0^2$. The slab is pulled in, so the agent does negative work and (C) is false.