In the given circuit, the current I through the battery will be

In the first circuit diagram, the diodes $D_1 \& D_3$ are forward biased and $D_2$ is reverse biased(Hence, no current through it). So the circuit is simplified to the second diagram.

Hence, the equivalent resistance is 

$R_{eq}=\frac{20\times 10}{20+10}$

$\Rightarrow R_{eq}=\frac{20}{3} \mathrm{\Omega }$

Using Ohm's law, the current is 

$I=\frac{10}{{\displaystyle \frac{20}{3}}}=\frac{30}{20} \mathrm{A}=1.5 \mathrm{A}$