In the given circuit diode $D_2$ is reverse biased, so it will not conduct. Diode $D_1$ and $D_3$ are forward biased, so they will conduct, Thus the equivalent circuit is



$\therefore R_{net}=\frac{(4+4)\times 16}{(4+4)+16}=\frac{8\times 16}{24}=\frac{16}{3} \mathrm{\Omega }$

$\therefore$ current through the battery,

$I=\frac{V}{R_{net}}=\frac{16}{\frac{16}{3}}=3 \mathrm{A}$