= Solution
{c}
For a <discrete valuation ring> $A$,
$$
\boxed{\dim A=1.}
$$
Let $\pi$ be a <uniformizer>, so its unique nonzero <maximal ideal> is $\mathfrak m=(\pi)$. We use the defining description of a <discrete valuation ring>: it is a non-field <local ring> that is a <principal ideal domain>, and every nonzero element is $u\pi^n$ for a <unit> $u$ and an integer $n\geq0$.
Every nonzero proper <ideal> is therefore $(\pi^n)$ for some $n\geq1$. Such an <ideal> is a <prime ideal> only when $n=1$: for $n\geq2$, $\pi\pi^{n-1}\in(\pi^n)$ while neither factor belongs to $(\pi^n)$. The <ideal> $(\pi)$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an <integral domain>. Thus
$$
\operatorname{Spec}A=\{(0),\mathfrak m\},\qquad (0)\subsetneq\mathfrak m,
$$
and the longest strict chain has length one.
The original PDF asks about a <discrete valuation ring>; the local TeX transcription's phrase “discrete variable $X$” is an extraction error.
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