A parallel plate capacitor with plate area A and plate separation d = 2   m has a capacitance of 4…

This parallel plate capacitor can be divided into two capacitors. One with dielectric $\left(C_1\right)$ and other without dielectric$\left(C_2\right)$. The two capacitors will be in series.

Initially $C=\frac{{\epsilon }_{0}A}{d}=4 \mathrm{\mu F}$

Now, $C_1=\frac{k{\epsilon }_{0}A}{{\displaystyle \frac{d}{2}}}=\frac{2\times 3\times {\epsilon }_{0}A}{{\displaystyle d}}=24 \mathrm{\mu F}$ and $C_2=\frac{{\epsilon }_{0}A}{{\displaystyle \frac{d}{2}}}=\frac{2\times {\epsilon }_{0}A}{{\displaystyle d}}=8 \mathrm{\mu F}$

Finally $C^{\text{'}}=\frac{C_1{C}_2}{C_1+C_2}=\frac{24\times 8}{24+8}=6 \mathrm{\mu F}$