Round-Robin fashion exercise
A round robin scheduler has a fifo queue of processes in state runnable.  At each time step the scheduler runs the process at the head of the queue for one quantum.

At the end of each time quantum the running process is interrupted and placed at the tail of the queue.  If the process blocks (requests i/o) during the time quantum it is typically completely removed from the queue.

The difference between your solution and your instructor's solution is the assumption you are making about what happens when a process's i/o request finishes, and the process again becomes runnable.   Your instructor assumes that the process is placed on the tail of the queue.  You seem to be assuming that the process is placed at the head of the queue.

Either choice (placing the newly unblocked process at the tail or at the head) is reasonable, but they have different consequences as described by @Gille's answer to a broader question.

If you place the newly unblocked process at the tail of the queue then you maintain the guarantee that no process will ever be starved.  In practice however, i/o-bound jobs are sometimes given higher priority, both to reduce the latency of those jobs, and to maximize throughput of i/o devices.  You could implement this increase in priority by putting newly unblocked processes at the head of the queue instead of the tail, but you would lose your starvation-freeness guarantees.