The junction diode shown in fig is ideal. The current in the circuit is

Let I be the current in the circuit,
As the diode is ideal so voltage drop across the diode is 0 since it is in forward bias.

4 V− 100I = 1 V

$\mathrm{I} = \frac{3}{100} = 0.03 \mathrm{A} = 30 \mathrm{mA}$