well, I don't know what to add more than saying that you can acheive the task by reading the packets in that order:
2, 1, 2, 5, 3, 4, 5 ; which is enough : no need to re-read every packet twice.
Then, with taht reading sequence (2, 1, 2, 5, 3, 4, 5) you have the cache-miss (« défaults de cache ») indicated in the solution.
Makes sense?