If the diodes $D_1$ and $D_2$ are ideal in given diagram. The current flowing in the circuit is

2A 4.59A 4.9A 3A

3A

D1 is in reversed biased hence no current will flow through it. $ I = \frac{24V}{3\Omega + 5\Omega} = 3A$ The correct answer is Option (4) → 3A