The answer is $\frac{1}{2}$.Note, that any triple $(a,b,c) = (1,2,c)$ satisfy the given inequality. So, $\frac{a}{b}$ takes the value
$\frac{1}{2}$. Now, we prove, that there is no other solution for $b \ge 2a$.
Let $t = |a^c-b!|$. If $t > 0$, $1 = \left | \frac{a^c}{t}-\frac{b!}{t} \right |$. Since $t \le b$, $\frac{b!}{t}$ is
an integer, so $\frac{a^c}{t}$ is also an integer. Furthermore, $2a \le b \Rightarrow a, 2a \in \left \{ 1,2,...,b \right \}$.
At least one of $a$ and $2a$ is different from $t$, so it is not canceled out from the product in
$\frac{b!}{t}$. So $a \: | \: \frac{b!}{t}$.
Therefore, since the difference between $\frac{b!}{t}$ and $\frac{a^c}{t}$ is $1$,
$gcd\left ( a,\frac{a^c}{t} \right )=1$ implying $t = a^c$. Thus, we are now left with two cases only: $\left | a^c-b! \right | = 0$ or $\left | a^c-b! \right | = a^c$. These cases reduce to: $b! = a^c$ and $b! = 2a^c$ respectively. Either way, $2a-1 \in \left \{ 1,2,...,b \right \}$,
so $ (2a-1)\: |\: b!\: |\: 2a^c$. Now, $gcd\left (2a-1, 2a^c \right )=1, \Rightarrow 2a-1= 1$ and $a = 1$.If $a = 1$, $b!-b \le 1 \Rightarrow b \le 2$. So, $(a,b) = (1,2)$ is the only solution at $b \ge 2a$. Done.
