The current (in A) in the network is 

Both diodes are in reverse biasing, so can be treated as infinite resistance or broken wires

Now all resistors are in series

$\begin{array}{l}\text{I}=\frac{V}{{\displaystyle R_{eff}}}=\frac{9}{(5+10+5+20)}=0.225\text{\hspace{0.17em}A}\\ \end{array}$