b
At steady state, current in the circuit is

\(i =\frac{4 V }{6+2+8}=\frac{1}{4}\,A\)

Voltage across \(C_1\) is

\(V _1= V _{ AC }=i(6 \Omega+2 \Omega)=\frac{1}{4} \times 8=2\,V\)

Voltage across \(C _2\) is

\(V _2= V _{ BD }=i(2 \Omega+8 \Omega)=\frac{1}{4} \times 10=2.5\,V\)

\(\Rightarrow \frac{ V _1}{ V _2}=\frac{2}{2.5}=\frac{4}{5}\)