Answer
(b)
Work Step by Step
According to the table the z-value for $90\%$ is: $z=1.64$.
Thus the confidence interval is: $\hat{p}\pm z\sqrt{\frac{p(1-p)}{n}}$, which here is: $0.8\pm 1.64\sqrt{\frac{0.8\cdot0.2}{4500}}=0.8\pm0.0098$, thus the answer is(b).