The ratio of ${\mathrm{Fe}}^{3+}\text{ and }{\mathrm{Fe}}^{2+}\text{ ions in }{\mathrm{Fe}}_{0.9}{ \mathrm{S}}_{1.0}$ is 

$$\begin{gathered} {\mathrm{Fe}}^{3+}=\mathrm{x} \\ {\mathrm{Fe}}^{2+}=0.9-\mathrm{x} \end{gathered}$$

Using charge balance 

$$\begin{gathered} {\mathrm{Fe}}_{0.9}{\mathrm{S}}_{1.0}=\mathrm{Charge} \mathrm{on} \mathrm{neutral} \mathrm{compound}=0 \\ 3\times \mathrm{x}2\times \left(0.93-\mathrm{x}\right)+2\times 1=0 \\ \mathrm{Charge} \mathrm{on} {\mathrm{Fe}}_{0.9} \\ \left({\mathrm{Fe}}^{+2} \& {\mathrm{Fe}}^{3+}\right) \\ \Rightarrow 3\mathrm{x}+2\left(0.9-\mathrm{x}\right)=2\times 2 \\ 3\mathrm{x}+1.8-2\mathrm{x}=2 \\ \mathrm{x}=2-1.8 \\ \mathrm{x}=0.2 \\ \mathrm{Thus}, \mathrm{the} \mathrm{no}. \mathrm{of} {\mathrm{Fe}}^{+2} \mathrm{ions}=0.93-\mathrm{x} \\ =0.93-0.2 \\ =0.73 \\ \frac{{\mathrm{Fe}}^{3+}}{{\mathrm{Fe}}^{2+}}=\frac{0.2}{0.73}=0.28 \end{gathered}$$