Cationic vacancies created when non-stoichiometric iron oxide is doped with silicon(IV) ions Problem

It is believed that non-stoichiometric compound $\ce{F_{0.93}O}$ forms by doping of $\ce{Fe^3+}$ ions in $\ce{FeO}$ crystal by replacement of $\ce{Fe^2+}$. Calculate the number of cationic vacancies if all of the $\ce{Fe^2+}$ ions are replaced by $\ce{Si^4+}$ ions in $\pu{0.1 mol}$ of $\ce{Fe_{0.93}O}$. Express your answer as a multiple of Avogadro's Number.

Answer stated by the book

 0.0465

My Solution
As the compound is electrically neutral I equated the net charges due to the cations and anions like this, assuming $x$ to be the fraction of $\ce{Fe^2+}$ in the compound.
$$[2x + 3(0.93-x)] - 2 = 0$$
This gave $x = 0.79$. Now, in $\pu{0.1 mol}$ of the compound there would be $\pu{0.079 mol}$ of $\ce{Fe^2+}$. As each silicon ion would replace two ions of $\ce{Fe^2+}$, one vacancy would be created for every replacement.
The number of replacements would be $\left(\frac{0.079}{2}\right)\times N_\mathrm{A} = 0.0395 N_\mathrm{A}$, and we would have as many vacancies. ($N_\mathrm{A}$ is Avogadro's Number)
Can anyone point out the error in my solution?
Your approach is correct, however, you forgot to account for the vacancies created when Fe3+ replaces Fe2+. First, calculate the cationic vacancies formed due to the replacement of Fe2+ by Fe3+.
Assuming  to be the fraction of Fe2+ in the compound, the fraction of Fe3+ in the compound is (0.93 - x).The total charge on x Fe2+ and (0.93 – x) Fe3+ should be equal to the charge due to O2- ions.
Hence, 2x + 3(0.93 – x) = 2
=&gt; 2x + 2.79 – 3x = 2.00 or x = 0.79. So, fraction of Fe3+ ions in the compound is 0.14
Now, 2 Fe3+ ions replace 3 Fe2+ ions, so there's one vacancy created for every 2 Fe3+ ions. That means for 0.1 mole of F0.93O, the number of vacancies will be 0.007A due to Fe3+ replacing Fe2+ in FeO. This is the part you missed.
Now, as you mentioned, each Si4+ ion will replace two Fe2+ ions in the compound, so again, one vacancy per replacement. And, as you already solved, The number of replacements would be (0.079/2)A = 0.0395A. So, the total number of vacancies are 0.0395NA + 0.007NA = 0.0465NA.
Hope this helps!