The cut-off voltage of the diodes (shown in figure) in forward bias is 0 . 6   V . The current through…

Here diode $D_1$ is forward biased and diode $D_2$ is reverse biased.

So, no current will flow through diode $D_2$ resistor.

Current through the resistor of $40 \mathrm{\Omega }$ is $i=\frac{1-0.6}{60+40}=\frac{0.4}{100}=4 \mathrm{mA}$