Answer
$\approx0.0308$
Work Step by Step
If $\hat{p}=0.5$
The z-value belonging to the $95\%$ confidence interval according to the table is $z=1.96$, thus the margin of error is: $z\sqrt{\frac{p(1-p)}{n}}$, which here is: $1.96\sqrt{\frac{0.5\cdot(1-0.5)}{1010}}\approx0.0308$
