Conditional Selection of Alphabet Letters

The problem is that you've double-counted some cases - the factor of 4 gives you 4 places where you put the A, but then the factor of $26^3$ includes the possibility that A is one of the other 3 letters, which means that patterns like ?A??A are being counted as both "A in 2nd place, 5th letter is free" and "A in 5th place, 2nd letter is free".

You might find it easier to take the difference of two counts, one of which is the number of words that don't have A anywhere.

ConMan's answer explains the OP's mistake. Further, as suggested by the answer of ConMan, the correct enumeration is $$25 \times \left[(26)^4 - (25)^4\right]. \tag1 $$

In (1) above, the 1st factor reflects the number of choices for the first letter. In the 2nd factor, the 1st term reflects all possible 4 character words, while the 2nd term reflects all possible 4 character words that do not use an "A".