This question is asking to calculate equilibrium constant for oxidation of glucose from thermodynamics quantities. So the first expression came to my mind was $-\Delta G^\circ=RT\ln K$. I found the value of $\Delta G^\circ=-2\,278.78$ but as I proceed I've got $\ln K=1\,161.94$ which I couldn't reach to the $K$ value. Is there any other ways to solve this question?
$$\ce{C6H12O6(s) + 6O2(g) <=> 6CO2(g) + 6H2O(l)}$$
$\Delta G^\circ_\mathrm r=-2878.78\ \mathrm{kJ}$
At equilibrium, $\Delta G=0$; $Q=K$,
$\Delta G=\Delta G^\circ+RT \ln K$
$2878.78\times10^3\ \mathrm J=\left(8.314\ \mathrm{J/(mol\ K)}\right) (298\ \mathrm K) \ln K$
$\ln K=1161.93596$
Answer from the book is $K=5\times10^{504}$