From the figure, It is clear that diodes $D_1$ and $D_3$ are reverse biased, therefore they will not conduct any current and the resistance of $11 \mathrm{\Omega }$ and $5 \mathrm{\Omega }$ will be not in use. The diode $D_2$ is forward biased, therefore the current flowing through the diode $D_2$ is same as the current through the battery.

On redrawing the circuit as shown, we can obtain the value of current as

$i=\frac{20}{7+3}=2 \mathrm{A}$