Answer
(c)
Work Step by Step
The critical value is given in the row $df=n-1=64-1=63\gt60$ and in the column with $c=0.9$ of the table. Thus $t_{\alpha/2}=1.671$.
Thus the confidence interval is: $\overline{x}\pm t_{\alpha/2}\frac{s}{\sqrt n}$, which here is: $15.5\pm1.671\frac{17.4}{\sqrt{64}}$, thus the interval is: $(11.87,19.13)$. Thus the answer is (c).