Answer is (a).
Given clock is $+$ edge triggered.
See the first positive edge. $X$ is $0$, and hence, the output is $0$. Q0 is $0$ and Q0' is $1$.
Second + edge, X is $1$ and Q0' is also one. So, output is 1. (When second positive edge of the clock arrives, Q0' would surely be $1$ because the setup time of flip-flop is given as $20$ ns and the clock period is $>= 40$ $ns$)
Third $+$ edge, $X$ is $1$ and Q0' is 0, So, output is 0. (Q0' becomes 0 before the 3rd positive edge, but output Y won't change as the flip-flop is positive edge triggered)
Now, output never changes back to 1 as Q0' is always 0 and when Q0' finally becomes 1, X is 0.
Set up time and hold times are given just to ensure that edge triggering works properly.