= Solution
{c}
Apply the <central limit theorem> to the <independent and identically distributed random variables> $X_i^2$, whose <mean> and <variance> are $\theta$ and $\theta^2$. It gives the approximate <standard normal distribution>
$$
\frac{\sqrt n(\widehat\theta-\theta)}{\theta}\ \approx\ N(0,1).
$$
The supplied <standard normal quantile> $c=1.96$ leaves <probability> $0.025$ in each tail. Hence
$$
\mathbb P\left(-1.96<\frac{\sqrt n(\widehat\theta-\theta)}{\theta}<1.96\right)\approx0.95.
$$
Since $\theta>0$, this event is equivalent to
$$
\theta\left(1-\frac{1.96}{\sqrt n}\right)<\widehat\theta<\theta\left(1+\frac{1.96}{\sqrt n}\right).
$$
For $n>1.96^2$, equivalently integer $n\geq4$, both denominators below are positive. Inverting the inequalities gives the approximate $95\%$ <confidence interval>
$$
\boxed{\left(\frac{\widehat\theta}{1+1.96/\sqrt n},\ \frac{\widehat\theta}{1-1.96/\sqrt n}\right),\qquad c=1.96.}
$$
The approximation is justified for large $n$; positivity at $n\geq4$ by itself does not guarantee accurate coverage. This construction is a <confidence interval from a relative normal pivot>: the unknown scale remains inside the pivotal ratio until the inequalities are inverted. Its meaning is that the random interval covers the fixed <statistical parameter> $\theta$ with <probability> approximately $0.95$ under repeated sampling.
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