Assume that we have 100 formula units of #"Fe"_0.93"O"_1.00#.
Then we have 93 #"Fe"# atoms and 100 #"O"# atoms.
The #"Fe"# atoms are a mix of #"Fe"^"2+"# and #"Fe"^"3+"# ions, and the #"O"# atoms are present as #"O"^"2-"# ions.
The total charge of the #"O"^"2-"# ions is 200-, so the total charge of the #"Fe"^"2+"# and #"Fe"^"3+"# ions must be 200+.
Let #x = "number of Fe"^"2+"color(white)(l) "ions"#.
Then #"93 – x" = "number of Fe"^"3+"color(white)(l) "ions"#.
The total + charge of these ions is
#2x + 3(93 - x) = 200#
#2x + 279 - 3x = 200#
#x = 279 - 200 = 79#
So, 100 formula units of #"Fe"_0.93"O"_1.00# contain
#"79 Fe"^"2+"#
#"14 Fe"^"3+"#
#"100 O"^"2-"#
#"% Fe"^"2+" = (79 × 55.84 color(red)(cancel(color(black)("u"))))/((79 × 55.84 + 14 × 55.84 + 100 ×16.00) color(red)(cancel(color(black)("u")))) × 100 % = 4411/6793 × 100 % = 64.93 %#
