Answer is (a).
Given clock is $+$ edge triggered.
See the first positive edge. $X$ is $0$, and hence, the output is $0$. $Q_0$ is $0$ and $Q_0'$ is $1$.
Second $+$ edge, $X$ is $1$ and $Q_0'$ is also $1.$ So, output is $1.$ (When second positive edge of the clock arrives, $Q_0'$ would surely be $1$ because the setup time of flip-flop is given as $20$ ns and the clock period is $\geq 40$ $ns$)
Third $+$ edge, $X$ is $1$ and $Q_0'$ is $0,$ So, output is $0.$ $(Q_0'$ becomes $0$ before the third positive edge, but output $Y$ would not change as the flip-flop is positive edge triggered)
Now, output never changes back to $1$ as $Q_0'$ is always $0$ and when $Q_0'$ finally becomes $1, X$ is $0.$
Set up time and hold times are given just to ensure that edge triggering works properly.