
This parallel plate capacitor can be divided into two capacitors. One with dielectric (C1) and other without dielectric(C2). The two capacitors will be in series.
Initially C=dϵ0A=4μF
Now, C1=2dkϵ0A=d2×3×ϵ0A=24μF and C2=2dϵ0A=d2×ϵ0A=8μF
Finally C′=C1+C2C1C2=24+824×8=6μF