My first answer was $5 \times 25 \times 24$, picking the vowel before the first and last letter. To my surprise, my book says this is the right answer! But I thought I was not counting everything and continued considering the following situations:
If I pick one vowel, there are $5 \times 21 \times 20$ words since I'm not considering vowels in the remaining choices. For instance, suppose I pick "abd". Then, there are $3!$ ways to arrange this word, but I want the middle letter to be the vowel, so only "bad" and "dab" are valid. Then for each of the $5 \times 21 \times 20$ words, there are two valid permutations. Therefore, there are $5 \times 21 \times 20 \times 2$ words.
If I pick two vowels, there are $5 \times 4 \times 21$ words. For instance, suppose I pick "aed". Again, there are $3!$ ways to arrange this word, but since there are two vowels, we have four valid permutations: "aed", "dea", "ead", and "dae". So for each of the $5 \times 4 \times 21$ words, there are four valid permutations. Therefore, there are $5 \times 4 \times 21 \times 4$ words.
If I pick three vowels, there are $5 \times 4 \times 3$ words. Since all are vowels, all $3!$ permutations are valid. Therefore, there are $5 \times 4 \times 3 \times 6$ words.
Finally, I summed all three to get $(5 \times 21 \times 20 \times 2) + (5 \times 4 \times 21 \times 4) + (5 \times 4 \times 3 \times 6)$ $3$-letter words with no repeated letters such that the middle letter is a vowel. If the answer in the book is correct, clearly I overcomplicated a simple problem, but I cannot see why the book is correct. Am I overcounting? Am I considering cases that should not be considered?
Thank you for any clarifications! :)