Solution:
To simplify notation, we will let $m$ be $m(U)$ and let $n$ be $|U|$. First note that there are 10 different diagonal-lengths in a regular 21-gon. Now consider the following set of 5 vertices.

Note that each of the 10 different diagonal-lengths appear (exactly once each). So for this set of 5 vertices we have $\frac{m}{n} = \frac{10}{5} = 2$. We will now show that this is the maximum possible value for $\frac{m}{n}$.

If $U$ is an arbitrary non-empty set of vertices, then there are two cases:

- Case 1: $n < 5$. The total number of pairs of vertices in $U$ is given by $\frac{1}{2} n(n - 1)$. Since $n - 1 < 4$ this gives us the bound:
$$m \leq \frac{n(n - 1)}{2} < \frac{n \times 4}{2} = 2n.$$
Thus $\frac{m}{n} < 2$ in this case.

- Case 2: $n \geq 5$. The total number of distances in $U$ is at most 10 because there are only 10 different diagonal lengths in the 21-gon. Therefore
$$\frac{m}{n} \leq \frac{10}{n} \leq \frac{10}{5} = 2$$
as required.