The $X$ at the beginning is fixed, and letters can be repeated, so we can just ignore it.
The vowel can be in one of $6$ places and there are $5$ possible vowels. The other $5$ letters can each be one of the $21$ consonants.
So in total there are $6\cdot 5 \cdot 21^5=122523030$ such words.
If the letters can't be repeated, consider the $5$ remaining letters:
The first one can be one of $20$ letters (not an $X$, not a vowel).
The second one can be one of $19$ letters (same as first, but can't equal the first either). etc.
So there are $6\cdot 5\cdot 20\cdot19\cdot 18\cdot 17\cdot 16=55814400$ words.