Solution:

A three-digit palindrome has the form $\overline{aba} = 101a + 10b$ with $a \in \{1, \ldots, 9\}$ and $b \in \{0, \ldots, 9\},$ and every such pair of digits occurs exactly once, so the two digits vary independently over the $90$ palindromes.

By linearity, the mean is $101$ times the average of $a$ plus $10$ times the average of $b,$ namely $$101 \cdot 5 + 10 \cdot \frac{9}{2} = 505 + 45 = 550.$$