In pure iron oxide $(FeO)$, iron and oxygen are present in the ratio $1: 1$.
However, here number of $Fe ^{2+}$ present $=0.93$
Or No. of $Fe ^{2+}$ ions missing $=0.07$
Since each $Fe ^{2+}$ ion has $2$ positive charge,
the total number of charge due to missing $( 0.07 ) Fe ^{2+}$ ions $=0.07 \times 2=0.14$
To maintain electrical neutrality, $0.14$ positive charge is compensated by the presence of $Fe ^{3+}$ ions.
Now since, replacement of one $Fe ^{2+}$ ion by one $Fe ^{3+}$ ion increases one positive charge, $0.14$ positive charge must be compensated by the presence of $0.14 \,Fe ^{3+}$ ions.
In short, $0.93\, Fe ^{2+}$ ions have $0.14\, Fe ^{3+}$ ions
$100 \,Fe ^{2+}$ ions have $=\frac{0.14}{0.93} \times 100=15.05 \%$