= Solution
A 95 percent <confidence set> is a measurable data-dependent set $C(X_1,\ldots,X_n)$ whose repeated-sampling coverage satisfies $\mathbb P_\theta\{\theta\in C(X_1,\ldots,X_n)\}\ge0.95$ for every parameter value, with equality for an exact 95 percent procedure. The parameter is fixed; the set is random. This is not a posterior probability statement about the parameter after the data have been observed.
For independent <normal random variables> of known variance, $\overline X\sim N(\mu,\sigma^2/n)$, so $(\overline X-\mu)/(\sigma/\sqrt n)$ has the standard <normal distribution>. The supplied quantile gives the exact <confidence interval>
$$
\boxed{\left[\overline X-1.960\frac\sigma{\sqrt n},\ \overline X+1.960\frac\sigma{\sqrt n}\right].}
$$
Its deterministic length is $3.920\sigma/\sqrt n$. Requiring this to be at most $\varepsilon$ gives
$$
\boxed{n\ge\frac{(3.920)^2\sigma^2}{\varepsilon^2},\qquad n_{\min}=\max\!\left(1,\left\lceil\frac{(3.920)^2\sigma^2}{\varepsilon^2}\right\rceil\right).}
$$
Among fixed-length intervals based equivariantly on the sample mean, the centred interval has greatest coverage: an interval of a given length captures most mass of a symmetric unimodal <normal distribution> when centred on its mean. Thus the width calculation is sharp for the usual fixed-length normal-mean construction.
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