Consider the problem
ceq=[(3μfp3μf)s(1μfp1μf)]p(1μf)=[(3+3)μfs(2μf)]p1μf=32+1=52μf
v=100vQ=cv=52×100=250μc
Charge stored across 1μf capacitor =100μc
ceq between A and B is 6μf=c
Potential drop across AB=v=Qc=25v
Potential drop across BC=75v
Hence,Option C is the correct answer.