In the given circuit, diode $D_1$ is reverse biased, so it will not conduct but $D_2,D_3$ are forward biased, so they will conduct, hence corresponding equivalent circuit. For the given circuit is

Now, the equivalent resistance of the circuit is

$R_{\mathrm{eq}}=\frac{\left(515\right)\times 20}{\left(5+5\right)+20}+\frac{20}{3} \mathrm{\Omega }$

Current through battery, $I=\frac{10 \mathrm{V}}{\frac{20}{3} \mathrm{\Omega }}=1.5 \mathrm{A}$