In the given figure, each diode has a forward bias resistance of...

We know that in reverse bias ideal diode does not pass current it only works in forward bias so, the current in the third branch is zero, it means the net resistance of the upper two branch is, $\dfrac{1}{{R}_{eq}} = \dfrac{1}{160} + \dfrac{1}{160} ⇒ {R}_{eq} = 80Ω$

$⇒ R{'}_{eq} = 80 + 20 = 100Ω$

$⇒ i = \dfrac{200}{100} = 2A$