If A is the area of each plate, the capacitance of the air-filled capacitor shown in Fig(a) is 

                   $C_0=\frac{{\epsilon }_{0}A}{d},$where C₀ = 2 µF.(given)

The capacitor in Fig.(b) will equivalent of a parallel combination of two capacitors 

The capacitance of air capacitor in Fig.(b) is

                   $C_1=\frac{{\epsilon }_{0}A/2}{d}=\frac{{\epsilon }_{0}A}{2d}=\frac{C_0}{2}$

The capacitance of dielectric filled capacitor in Fig.(b) is 

                  $C_2=\frac{k{\epsilon }_{0}A/2}{d}=\frac{k{\epsilon }_{0}A}{2d}=\frac{k{C}_0}{2}$

Since C₁ and C₂ are in parallel, the capacitance C of the capacitor shown in Fig.(b) is

                $$\begin{gathered} C=C_1+C_2=\frac{C_0}{2}+\frac{k{C}_0}{2} \\ =\frac{C_0}{2}\left(1+k\right)=\frac{2\mu F}{2}\left(1+3\right)=4\mu F \end{gathered}$$