Determining the temperature at which 90% of the reactants in a equilibrium have reacted
Assuming $\ce{FeO}$ and $\ce{Fe3O4}$ are solid throughout the range of temperatures we are interested in, they won't appear in the expression for the equilibrium constant $K$ (which in this case is equivalent to $K_p$). If $n$ is the total number of moles of $\ce{CO}$ that you start with,

$$K_p = K = \frac{p_{\ce{CO2}}}{p_{\ce{CO}}} = \frac{n_{\ce{CO2}}}{n_{\ce{CO}}} = \frac{0.9 n}{0.1 n} = 9$$

Plugging this into the formula previously derived, we find $T = 1388~\mathrm{K}$.