Answer
See the detailed answer below.
Work Step by Step
The dielectric material here doesn't fill the whole space between the two plates as we see below.
We know that the potential difference inside a vacuum parallel plate capacitor is given by $V=E_0 s$ where $s$ is measured from the negative plate.
And hence,
$$\Delta V_C=E_0d\tag 1$$
So, in this case, after adding the dielectric material, as we see in the figure below.
$$(\Delta V_C)_{new}= \int_{0}^dE_0ds$$
we have here 3 areas, starting from the negative plate,
$$(\Delta V_C)_{new}= \int_{0}^dE_0ds=\dfrac{E_0d}{4}+\dfrac{E_0d}{2\kappa}+\dfrac{E_0d}{4}$$
Thus,
$$(\Delta V_C)_{new}= E_0d\left[ \dfrac{1}{4}+\dfrac{1}{2\kappa}+\dfrac{1}{4}\right] $$
$$(\Delta V_C)_{new}=E_0d\left[ \dfrac{\kappa+1}{2\kappa} \right]\tag 1$$
Recalling that the capacitance is given by
$$C_{\rm new}=\dfrac{Q}{(\Delta V_C)_{new}}$$
Plug from (1),
$$\boxed{C_{\rm new}=\dfrac{2\kappa Q}{ E_0d(\kappa+1)}=\dfrac{2\kappa C_0}{\kappa+1}}$$
