Your mistake is in the arithmetic. What you think comes out to 2341 really does come out to 240.
$4^5=1024$, $3^5=243$, $2^5=32$, $1024-(4)(243)+(6)(32)-4=1024-972+192-4=1216-976=240$
Let $o(n)$ be the number of such words of length $n$ with an odd number of $A$’s. We can build these words by choosing an odd $k\in\{0,1,\ldots,n\}$, picking $k$ positions in the word for $A$’s, and filling the other $n-k$ positions with letters from the set $\{B,C,D,E,F\}$; given $k$, this can be done in $\binom{n}k5^{n-k}$ ways, so
$$o(n)=\sum_{{k=0}\atop{k\text{ odd}}}^n\binom{n}k5^{n-k}\;.$$
Let $e(n)$ be the number of words of length $n$ with an even number of $A$’s; clearly
$$e(n)=\sum_{{k=0}\atop{k\text{ even}}}^n\binom{n}k5^{n-k}\;.$$
There are $6^n$ words of length $n$, so $$e(n)+o(n)=6^n\;.\tag{1}$$ On the other hand,
$$\begin{align*}
e(n)-o(n)&=\sum_{{k=0}\atop{k\text{ even}}}^n\binom{n}k5^{n-k}-\sum_{{k=0}\atop{k\text{ odd}}}^n\binom{n}k5^{n-k}\\\\
&=\sum_{k=0}^n\binom{n}k(-1)^k5^{n-k}\\\\
&=(-1+5)^n\\\\
&=4^n\;.
\end{align*}\tag{2}$$
Combine $(1)$ and $(2)$ to solve for $o(n)$ and $e(n)$.
Added: You can also approach it by finding a recurrence and solving it for a closed form. Let $O_n$ be the set of words of length $n$ with an odd number of $A$’s, and let $E_n$ be the set of words of length $n$ with an even number of $A$’s. Let $w\in O_n$, let $w_f$ and $w_\ell$ be respectively the first and last letters of $w$, and let $w'$ be the word of length $n-2$ obtained by deleting $w_f$ and $w_\ell$ from $w$. If exactly one of $w_f$ and $w_\ell$ is an $A$, then $w'\in E_{n-2}$; otherwise, $w'\in O_{n-2}$. Conversely, any $w'\in E_{n-2}$ can be extended to a word in $O_n$ in $10$ ways by adding an $A$ at one end and a non-$A$ at the other, and any $w'\in O_{n-2}$ can be extended to a word in $O_n$ in $26$ ways by appending an $A$ at each end or a non-$A$ at each end. It follows that
$$\begin{align*}
o(n)&=10e(n-2)+26o(n-2)\\
&=10\big(o(n-2)+e(n-2)\big)+16o(n-2)\\
&=10\cdot6^{n-2}+16o(n-2)\;.
\end{align*}$$
This recurrence can be solved by a variety of techniques.
Best Answer
Hint: How many four letter words can you make if you include one B? How many four letter words can you make if you include two B's? How many four letter words can you make if you include three B's?
Good luck!