The circuit shown in the figure contains two diodes each with a forward resistance of 30 Ω and with infinite backward resistance. If the battery is 3 V, the current through the 50 Ω resistance is:

0.02 A

In the circuit the upper diode $D_1$ is reverse biased and the lower diode $D_2$ is forward biased. Thus there will be no current across upper diode junction. The effective circuit will be as shown in figure.

Total resistance of circuit

$R = 50Ω + 70Ω + 30Ω = 150Ω$

Current in circuit, $I=\frac{V}{R}=\frac{3V}{150Ω}=0.02A$