Ratio of MbO2/Mb in a satured solution of dissolved oxygen from dry air


> 
> What is the ratio of $\ce{MbO2/Mb}$ if the solution is saturated with air? 
> 
> 
> **Details of the question**:
>  $$
> \begin{align}
> \ce{Mb + O2 (aq) &-> MbO2}\\
> \Delta G' &= -30.3~\mathrm{kJ/mol}\\
> T &= 298~\mathrm{K}\\
> \ce{pH} &= 7\\
> \text{Henry's Constant for }\ce{O2} &= 42 \cdot 10^3~\mathrm{bar}\\
> \text{Content of }\ce{O2}\text{ in dry air} &= 21\%\\
> \end{align}
> $$
> 
> 
> 


**Attempt at Solution**:


So, basically this is a two part question, as I see it. First I need to figure out the concentration of dissolved $\ce{O2}$ using Henry's Law. Then I can simply plug this value into the $$\Delta G' = -RT\ln K$$ equation where $$K = \frac{\ce{[MbO2]}}{\ce{[O2][Mb]}}.$$


Therefore:
$$
\begin{align}
P &= K\cdot m\\
0.21~\mathrm{bar} &= (42 \cdot 10^3)\cdot m\\
m &= 5 \cdot 10^{-6} \pu{mol/kg}\\
-30.300~\mathrm{J/mol} &= -8.314 \cdot 298 \cdot \ln\left(\frac{\ce{[MbO2]}}{[5\cdot10^{-6}]\ce{[Mb]}}\right)
\end{align}
$$
After dividing and raising both sides to $e^x$ (to get rid of $\ln$) I get $$\frac{\ce{MbO2}}{\ce{Mb}} = 1.02.$$


However the answer sheet says it is $21.6$. I cannot figure out why this is so. 


