A composite parallel plate capacitor is made up of two different...

Capacitance of each capacitor

${C}_{1} = \dfrac{A3{ε}_{0}}{\dfrac{1}{2}} = 6A{ϵ}_{0}$ and ${C}_{2} = \dfrac{A4{ϵ}_{0}}{1} = 4A{ε}_{0}$

Here, $A$ is area of plate.

Equivalent capacitance is

${C}_{eq} = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}} = \dfrac{6A{ε}_{0} \times 4A{ε}_{0}}{6A{ε}_{0} + 4A{ε}_{0}} = \dfrac{24}{10}A{ϵ}_{0}$

Now, net charge stored is ${q}_{net} = {C}_{eq}\left(\Delta V\right) = \dfrac{24}{10}A{ε}_{0}\left(100\right) = 240A{ε}_{0}$

Potential drop across ${C}_{2}$ is $\Delta{V}_{2} = \dfrac{240A{ε}_{0}}{4A{ε}_{0}} = 60V$

Thus, voltage of the conducting foil is ${V}_{foil} = 60V$