Answer
(a).
Work Step by Step
$z=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{172-188}{41/\sqrt{56}}\approx-2.92$
Thus, using the table, the probability of obtaining a result more extreme than $z$: $P(Z\lt-2.92)=0.0018$.
If $P\lt\alpha$, then the null hypothesis is rejected and is statistically significant at $\alpha$. $0.0018\lt0.01$, thus the answer is (a).