Let us trace the Round Robin execution with a time quantum of 50 ms:
t=0 ms: Process A arrives with CPU burst 100 ms. Ready Queue: [A]. Process A executes for its time slice from t=0 to 50.
During this time, Process B arrives at t=5 ms and Process C arrives at t=10 ms.
t=50 ms: Process A's time slice expires (A has 50 ms CPU remaining). A is preempted and added to the tail of the queue. Ready Queue: [B,C,A].
t=50 to 100 ms: Process B executes for 50 ms (B has 300 ms remaining). Ready Queue: [C,A,B].
t=100 to 150 ms: Process C executes for 50 ms (C has 150 ms remaining). Ready Queue: [A,B,C].
t=150 to 200 ms: Process A executes its remaining 50 ms CPU burst and completes its CPU phase for iteration 1. A immediately moves to I/O for 500 ms (from t=200 to 700 ms). Ready Queue: [B,C].
t=200 to 250 ms: Process B executes for 50 ms (B has 250 ms remaining). Ready Queue: [C,B].
t=250 to 300 ms: Process C executes for 50 ms (C has 100 ms remaining). Ready Queue: [B,C].
t=300 to 350 ms: Process B executes for 50 ms (B has 200 ms remaining). Ready Queue: [C,B].
t=350 to 400 ms: Process C executes for 50 ms (C has 50 ms remaining). Ready Queue: [B,C].
t=400 to 450 ms: Process B executes for 50 ms (B has 150 ms remaining). Ready Queue: [C,B].
t=450 to 500 ms: Process C executes its remaining 50 ms CPU burst and completes its CPU execution at t=500 ms.
At t=500 ms, Process C initiates its first I/O operation of duration tio=500 ms.
Therefore, Process C completes its first I/O operation at:
500+500=1000 ms