Answer:
The answer is 500 kbps
Explanation:
Consider the given data in the question.
R1 = 500 kbps
R2=2 Mbps
R3 = 1 Mbps
Now as it is mentioned that there is no other traffic in the network.
Thus,
throughput of the file = min {R1,R2,R3}
throughput of the file = min {500 kbps, 2 Mbps, 1 Mbps}
T/P of the file = 500 kbps