Confidence interval for the square of binomial probability

I have a binomial distribution where the estimate for p is 0.03 out of 1000 sample trials.

Using the normal approximation and Chi-square distribution for the square of normal distribution, how can I find a confidence interval for the square of p?

$$\hat p \sim N\left(p, \frac {p(1-p)}n\right)$$ $$\frac {\sqrt n(\hat p -p)}{\sqrt{p(1-p)}} \sim N\left(0,1\right)$$ $$\frac {n(\hat p -p)^2}{p(1-p)} \sim \chi^2(df=1)$$
Here $$\sim$$ means "asymptotically following"
Find the percentile of $$\chi^2$$ distribution. Let $$X$$ be its 2.5 percentile. $$X<\frac {n(\hat p -p)^2}{p(1-p)}$$ $$-(X+n)p^2 + (X+2n\hat p)p -n\hat p^2 <0$$ You can get the solution following the quadratic formula. Similarly, you can get other one using 97.5% percentile.
There are better methods for your case: First, there are exact methods for a confidence interval for a binomial $$p$$, see the many relevant posts here.
Use one such method and get a confidence interval $$(l, u)$$ for $$p$$. Then $$(l^2, u^2)$$ is a confidence interval for $$p^2$$.