The answer that you suggest is not right, unfortunately in more than one way. Here is a concrete analysis of the problem, lengthy but I hope fairly straightforward. Your mistake was to try to use a simple formula, before carrying out an analysis of what is really going on.
How many $11$ letter words have FRED as a subword, with the letters FRED, in that order, occupying the first $4$ positions in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$.
How many words have FRED as a subword, with the letters FRED, in that order, occupying positions $2$ to $5$ in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$.
The same is true for the words that have FRED occupying positions $3$ to $6$, $4$ to $7$, $5$ to $8$, and so on up to $8$ to $11$.
Now it is tempting to add up, getting a total of $(8)(26^7)$. However, this will double count, for example, the words that start with FRED, then some useless letter, then FRED again, with $2$ letters added at the end. There are several other types of words that have two FRED's in them that will be double-counted.
Luckily, we don't have to worry about words that have $3$ or more occurrences of FRED. (If we were dealing with $47$-letter words, things could get very unpleasant.)
Let us find out how many double-counted words there are. We will do it slowly, and then in a slicker way.
How many "two FRED" words have the first one at the beginning? The second one can then start in any of positions $5$ to $8$, so $4$ positions in all. For each of thse choices, the remaining "useless" letters can be placed in $26^3$ ways, for a total of $(4)(26^3)$.
How many "two FRED" words have the first FRED starting at position $2$? The same reasoning as in the paragraph above shows that there are $(3)(26^3)$ such words.
How many have the first FRED starting at position $3$? Clearly $(2)(26^3)$. How many with the first FRED starting at position $4$? Clearly $(1)(26^3)$. And that's all.
So in total the number of "two FRED" words is $(10)(26^3)$.
Recall that with some double-counting, we got a total of $(8)(26^7)$. Get rid of the double counting by subtracting $(10)(26^3)$ and we get $(8)(26^7) -(10)(26^3)$
A slicker way: We describe another way of counting the "two FRED" words. Take some Scrabble tiles, and glue them together to form two copies of the word "FRED." Think of these as two special mega letters. To make a "two FRED" word, we take our $2$ mega letters, and $3$ ordinary letter, and put them in a row. So we are making a weird $5$-letter word. The positions of the two mega letters can be chosen in $\binom{5}{2}$ ways. For each such choice, the other letters can be filled in $26^3$ ways, for a total of $\binom{5}{2}(26^3)$. This is exactly the count of $(10)(26^3)$ that we got in a slower way earlier.
Cute, but maybe more than a little dangerous. The more plodding approach probably lets us retain greater control over what is going on.
Comment: It is all too easy to count in a plausible way that turns out to be incorrect. As a partial check, one might try to use the same reasoning on say $5$-letter words, and MO instead of FRED. Use the kind of "general" reasoning mentioned above, and compare with a detailed hand-count.