First $P_0$ will enter the while loop as $S_0$ is $1.$ Now, it releases both $S_1$ and $S_2$ and one of them must execute next. Let that be $P_1.$ Now, $P_0$ will be waiting for $P_1$ to finish. But in the mean time $P_2$ can also start execution. So, there is a chance that before $P_0$ enters the second iteration both $P_1$ and $P_2$ would have done release $(S_0)$ which would make $S_1$ $1$ only (as it is a binary semaphore). So, $P_0$ can do only one more iteration printing $'0'$ two times.
If $P_2$ does release $(S_0)$ only after $P_0$ starts its second iteration, then $P_0$ would do three iterations printing $'0'$ three times.
If the semaphore had $3$ values possible (an integer semaphore and not a binary one), exactly three $'0's$ would have been printed.
Correct Answer: A, at least twice