The current $$i$$ in the network is:

The top-left diode is forward-biased and acts as a short circuit.

The bottom-right diode is reverse-biased and acts as an open circuit, completely disconnecting its branch.

The circuit reduces to a single continuous loop where all active resistors are connected in series:

$$R_{\text{total}} = 5\ \Omega + 10\ \Omega + 5\ \Omega + 10\ \Omega = 30\ \Omega$$

$$i = \frac{9}{(5 + 10 + 5 + 10)} = \frac{9}{30}\text{ A} = 0.3\text{ A}$$