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