The current i in the network is:

(3) Both the diodes are reverse biased, so, there is no flow of current through $5 \Omega$ and $20 \Omega$ resistances.

Now, two resistors of $10 \Omega$ and two resistors of $5 \Omega$ are in series.

Hence current $I$ through the network

$I=\frac{V}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{10+5+10+5}$

$\Rightarrow I=\frac{9}{30}=0.3 \mathrm{~A}$