Answer
(b)
Work Step by Step
The critical value is given in the row $df=n-1=43-1=42\gt40$ and in the column with $c=0.95$ of the table. Thus $t_{\alpha/2}=2.021$.
Thus the confidence interval is: $\overline{x}\pm t_{\alpha/2}\frac{s}{\sqrt n}$, which here is: $3.6\pm2.021\frac{1.26}{\sqrt{43}}=3.6\pm0.39$, thus the interval is: $(3.21,3.99)$. Thus the answer is (b).