Let, the number of $M ^{2+}$ ions $= x$
Then, the number of $M ^{3+}$ ions will be $0.96- x$
We know, the overall charge in the metal oxide is zero.
So, $x(2)+(0.96-x)(3)+1(-2)=0$
$\Rightarrow 2 x+2.88-3 x=2$
$\Rightarrow- x =-0.88$
$\Rightarrow x =0.88$
$\therefore$ Number of $M ^{3+}$ ions $=0.96-0.88=0.08$
$\therefore$ Percentage of $M ^{3+}$ ions $=0.08 / 0.96 \times 100=8.33 \%$