1. Interpret the formula

The oxide has formula $M_{0.98}O$.

This means for every $1$ oxygen atom, there are $0.98$ metal atoms.

Since oxygen is present as $O^{2-}$, total negative charge per formula unit is:

$-2$

So, total positive charge from all metal ions together must be:

$+2$

2. Let the fraction of metal atoms present as $M^{3+}$ be $x$

Then fraction present as $M^{2+}$ is:

$1-x$

Average oxidation state of metal $M$ will be:

$2(1-x) + 3x = 2 + x$

3. Use charge neutrality

There are $0.98$ metal atoms per oxygen atom.

Hence total positive charge contributed by metal ions is:

$0.98(2+x)$

This must balance the $2-$ charge of oxygen:

$0.98(2+x)=2$

Solve:

$2+x = \frac{2}{0.98} = \frac{100}{49}$

So,

$x = \frac{100}{49}-2 = \frac{100-98}{49} = \frac{2}{49}$

$x \approx 0.0408$

Thus, fraction of metal as $M^{3+}$ is:

$0.0408 \times 100 = 4.08\%$

4. Check options

• A: $7.01\%$ ❌
• B: $4.08\%$ ✅
• C: $6.05\%$ ❌
• D: $5.08\%$ ❌

5. Final Answer

The fraction of metal existing as $M^{3+}$ is:

$\boxed{4.08\%}$