$C _{1}=\frac{ A 3 \in_{0}}{\frac{1}{2}}=6 A \in_{0}$

$C _{2}= A 4 E _{0}=4 A \in_{0}$

Equivalent capacitance

$C _{\text {eq }}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}} \Rightarrow \frac{24}{10} A \in_{0}$

$q _{\text {net }}= C _{\text {eq }}(\Delta V ) \Rightarrow 240 A \in_{0}$

$\Delta V _{2}=\frac{240 A \in_{0}}{4 A \in_{0}}=60\,V$

$\left(\Delta V _{2}=\text { Potential drop across } C _{2}\right)$

$V _{\text {foil }}=60\,V$