Let, the number of M2+ ions =x
Then, the number of M3+ 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⇒2x+2.88−3x=2⇒−x=−0.88⇒x=0.88∴ Number of M3+ ions =0.96−0.88=0.08∴ Percentage of M3+ ions =0.08/0.96×100=8.33%