The circuit contains two diodes each with a forward resistance of...

In this circuit ${D}_{1}$ will be forward bias and ${D}_{2}$ will be revers bias.

$∴$ There will be no current through ${D}_{2}$ and ${R}_{2}$

Apply KVL in circuit we get

$+ 6 - 50i - 130i - 120i = 0$

$i = \dfrac{6}{300}A = \dfrac{6}{300} \times 1000mA$

$⇒ 20mA$