The given data is
A=200×10−4m2d=5×10−3m
The value of a+b=4×10−3m
The formula for a parallel plate capacitor is
C=Dϵ0A
Let the capacitor with gap a be C1 and capacitor with gap b be C2
C1=aϵ0A,C2=bϵ0A
The equivalent capacitance is
Ceq1=ϵ0Aa+ϵ0Ab⇒Ceq=a+bϵ0A
It is given that Ceq=xϵ0F
So,
a+bϵ0A=xϵ0⇒x=a+bA⇒x=4×10−3200×10−4=5