I think I am stuck in front of a very standard exercise of statistics, thus an operative hint on what I should be doing would be considered a sufficient answer.
Define $$\mathcal{A}_n = \frac{1}{n} \sum_1^n X_i^2,$$ I am requested to use the $\mathsf{CLT}$ and the fact that $q_{0,05}(\mathcal{A}_1)=3,84$ to approximate the $0.95$-quantile $q_{0,05}(\mathcal{A}_n)$ for the sample size $n=10^2$.
My considerations up to this point have been:
- $\mathcal{A_1} \sim \chi^2_1$
- $\sqrt{n}(\mathcal{A_n}-1) \to \mathcal{N}(0,1)$
- $n \cdot\mathcal{A_n} \sim \chi^2_n$
I think $\sqrt{n}(\mathcal{A_n}-1) \to \mathcal{N}(0,2)$ or $\sqrt{\frac n2}(\mathcal{A_n}-1) \to \mathcal{N}(0,1)$
I also suspect it is $q_{0.95}(\mathcal{A}_1)=3.84 \approx 1.96^2$
suggesting for a standard normal random variable $X \sim \mathcal{N}(0,1)$ you have $q_{0.975}(X)\approx 1.96$ (taking account of the two tails),
but it could be more useful to know $q_{0.90}(\mathcal{A}_1)\approx 2.706\approx 1.645^2$
and so $q_{0.95}(X)\approx 1.645$.
Putting those together suggests $q_{0.95}\left(\sqrt{\frac {100}2}(\mathcal{A}_{100}-1)\right) \approx 1.645$
and so $q_{0.95}\left(\mathcal{A}_{100}\right) \approx 1+ \frac{1.645}{\sqrt{50}} \approx 1.23$. The exact answer seems to be closer to $1.24$