The expected length of the constructed confidence interval is
$2n\theta (\frac{1}{\chi^2_{2n, 0.975}} - \frac{1}{\chi^2_{2n, 0.025}})$ goes to 0 as n goes to infinity.
I don't know why the above is true. $\chi^2_{2n, 0.025}$ means the 97.5% quantile. (0.025 is upper tail.)
The solution is below. I don't know where $2n+1.96\sqrt{4n}$ come from.
