Determining Kc for the equilibrium involving iron thiocyanate using spectrophotometric data $$\ce{Fe^3+(aq) + SCN-(aq) &lt;=&gt; [Fe(SCN)]^2+(aq)}$$
$\pu{12.0 mL}$ of $\pu{0.00110 mol L-1}$ $\ce{Fe^3+}$ was added to $\pu{6.00 mL}$ of $\pu{0.00140 mol L-1}$ $\ce{SCN-}$. The absorbance of the resulting solution was $0.625$. If the constant $(\varepsilon b) = \pu{3800 (mol L-1)-1}$, determine the value of $K_c$ that should be reported for the equilibrium.
a. $\pu{1.64e-4}$
b. $\pu{3.03e-4}$
c. $\pu{4.67e-4}$
d. $\pu{5.69e-4}$
e. $957$

Answer given: e.
Given what I know about $K_c$ it should be:
$$K_c = \frac{\left[\ce{[Fe(SCN)]^2+}\right]}{[\ce{Fe^3+}][\ce{SCN-}]}$$
which gives me the answer a. I tried calculating $C$ by manipulating Beer's law, but I still don't end up with $957$ no matter what I try.
I would appreciate it if someone could walk me through the steps of solving this problem.
The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as:
$$A = \epsilon bc$$

where:


A=absorbance
$\epsilon$ = molar attenuation coefficient
$b$ = path length
$c$ = concentration


So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is:

$$ c = \dfrac{A}{\epsilon b} = \dfrac{0.625}{3800} = 1.64\times10^{-4}$$

Considering the dilutions...

$\ce{[Fe^{+3}] + [Fe(SCN)]^2+} = 0.00110 \times \dfrac{12}{18} = 7.33\times10^{-4}$

$\ce{[SCN^-] + [Fe(SCN)]^2+} = 0.00140 \times \dfrac{6}{18} = 4.67\times10^{-4}$

But there is $1.64\times10^{-4}$ of $\ce{[Fe(SCN)]^2+}$ so:

$\ce{[Fe^{+3}]} = 7.33\times10^{-4} - 1.64\times10^{-4} = 5.69\times10^{-4}$

$\ce{[SCN^-]} = 4.67\times10^{-4} - 1.64\times10^{-4} = 3.03\times10^{-4}$

Therefore:

$K_c = \dfrac{1.64\times10^{-4}}{(5.69\times10^{-4})(3.03\times10^{-4})} = 951$

I'd assume the small difference is just due to a difference in rounding somewhere...