= Solution
A $100(1-\alpha)\%$ <confidence interval> procedure produces random endpoints $L,U$ such that $\mathbb P_\mu(L\leq\mu\leq U)=1-\alpha$, under the model. In everyday terms, repeating the same sampling and interval construction many times would give intervals containing the true mean in about $100(1-\alpha)\%$ of repetitions.
For $n\geq2$, define the <sample mean> $\bar X=n^{-1}\sum_iX_i$ and the <sample variance> $S^2=(n-1)^{-1}\sum_i(X_i-\bar X)^2$. For <independent> observations from a <normal distribution> with $\sigma^2>0$, the standard sampling results are
$$
\frac{\sqrt n(\bar X-\mu)}S\sim t_{n-1},\qquad\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}.
$$
The first follows from the <independent> <normal sample> mean and <sample variance>. Let $q=t_{n-1,1-\alpha/2}$ be the positive Student <quantile>. Symmetry gives the central <confidence interval>
$$
\boxed{\left[\bar X-q\frac S{\sqrt n},\ \bar X+q\frac S{\sqrt n}\right].}
$$
Its width is $W=2qS/\sqrt n$, so
$$
\boxed{c=\frac{n(n-1)}{4t_{n-1,1-\alpha/2}^{\,2}},\qquad \frac{cW^2}{\sigma^2}=\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}.}
$$
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