The formula of a metal oxide is ${\mathrm{M}}_{0.96}{\mathrm{O}}_1$. The fractions of metal that exists as ${\mathrm{M}}^{3+}$and ${\mathrm{M}}^{2+}$ ions in that oxide are respectively

${\mathrm{M}}_{0.96}{\mathrm{O}}_1$ means that ratio of M:O is $\frac{0.96}{1.00} \mathrm{or} \frac{96}{100}.$ These means that for 100 oxygen atoms, there are 96 M atoms.

Each oxygen has -2 charge, this means that 100 atoms has -200 charge. 

Assume that the number of M²⁺ atoms is x, then M³⁺ atoms are (0.96-x). Total charge of M is:

$\left(+2\right)\mathrm{x}+\left(+3\right)\left(0.96-\mathrm{x}\right)=(288-\mathrm{x})$

Now, the compound is electrically neutral. This means that charge of M and O are same.

 $$\begin{gathered} 288-\mathrm{x}=200 \\ \mathrm{x}=88 \mathrm{or} 0.88 \\ 0.88 \mathrm{for} {\mathrm{M}}^{2+} \\ 0.08 \mathrm{for} {\mathrm{M}}^{3+} \end{gathered}$$