Answer
See below.
Work Step by Step
$np=6000\cdot0.9=5400\gt10$, $n(1-p)=6000\cdot0.1=600\gt10$, thus the sampling distribution of $\hat{p}$ is approximately normal with a mean of $p=0.9$ and a standard deviation of $\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.9\cdot0.1}{6000}}\approx0.0039$