Charge on cap $=1 \mathrm{mC}$
$$
\mathrm{V}_{\text {capacitor }}=\frac{Q}{C}=\frac{1 \times 10^{-3}}{5 \times 10^{-6}}=\frac{1000}{5}=200 \mathrm{~V}
$$
So, current through $R_5$
$$
I=\frac{V}{R}=\frac{200}{10}=20 \mathrm{~A}
$$
Hence, we have following current distribution.

In loop $B C D B$,

In loop $A B C D A$,
$$
\begin{aligned}
& -200-15 R_3+50+250=0 \\
& 15 R_3=100 \Rightarrow R_3=\frac{100}{15} \Omega
\end{aligned}
$$
From Eq. (i), we get
$$
5 R_2=50 \Rightarrow R_2=10 \Omega
$$
and from loop $A D C A$,
$$
\begin{aligned}
& -250-50+310-25 R_1=0 \\
& 25 R_1=10 \Rightarrow R_1=\frac{10}{25} \Omega
\end{aligned}
$$
So, $\frac{R_1 \times R_2}{R_3}=\frac{\frac{10}{25} \times 10}{\left(\frac{100}{15}\right)}=\frac{15}{25}=\frac{3}{5}=0.6$