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

In the first circuit diagram, the diodes ${D}_{1} \text{ and } {D}_{3}$ 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

${R}_{eq} = \dfrac{20 \times 10}{20 + 10}$

$⇒ {R}_{eq} = \dfrac{20}{3}Ω$

Using Ohm's law, the current is

$I = \dfrac{10}{\dfrac{20}{3}} = \dfrac{30}{20}A = 1 . 5A$