Experimentally it was found that a metal oxide has formula $$M_{0.98}O$$. Metal M, is present as $$M^{2+}$$ and $$M^{3+}$$ in its oxide. Fraction of the metal which exists as $$M^{3+}$$ would be:

We are told that the empirical formula of the oxide is $$M_{0.98}O$$. This means that, for every one oxide ion $$O^{2-}$$ present, there are only $$0.98$$ atoms of the metal $$M$$ in the solid.

The metal occurs in two oxidation states, $$M^{2+}$$ and $$M^{3+}$$. Let us assume that a fraction $$x$$ (in decimal form) of the total metal atoms is present as $$M^{3+}$$. Consequently, the remaining fraction $$1 - x$$ will be present as $$M^{2+}$$.

So, out of the total $$0.98$$ metal atoms,

$$ \text{Number of }M^{3+}\text{ ions} = 0.98\,x, \qquad \text{Number of }M^{2+}\text{ ions} = 0.98\,(1 - x). $$

Next, we impose the requirement of overall electrical neutrality. The oxide ion carries a charge of $$-2$$. Therefore, the sum of positive charges provided by the metal ions must equal $$+2$$.

First, we compute the positive charge contributed by each type of metal ion.

Charge from $$M^{2+}: \quad 2 \times 0.98\,(1 - x) = 1.96\,(1 - x).$$

Charge from $$M^{3+}: \quad 3 \times 0.98\,x = 2.94\,x.$$

The total positive charge is the sum of these two:

$$ \text{Total positive charge} = 1.96\,(1 - x) + 2.94\,x. $$

We now set this equal to the required $$+2$$ charge from one $$O^{2-}$$:

$$ 1.96\,(1 - x) + 2.94\,x \;=\; 2. $$

Let us expand and simplify step by step.

First, expand the term $$1.96\,(1 - x)$$:

$$ 1.96\,(1 - x) = 1.96 - 1.96\,x. $$

Add the other term $$2.94\,x$$:

$$ 1.96 - 1.96\,x + 2.94\,x \;=\; 2. $$

Combine the $$x$$ terms:

$$ 1.96 + (2.94 - 1.96)\,x = 2. $$

Calculate the coefficient of $$x$$ in parentheses:

$$ 2.94 - 1.96 = 0.98, $$

so the equation becomes

$$ 1.96 + 0.98\,x = 2. $$

Now isolate $$x$$ by subtracting $$1.96$$ from both sides:

$$ 0.98\,x = 2 - 1.96 = 0.04. $$

Finally, divide by $$0.98$$ to solve for $$x$$:

$$ x = \frac{0.04}{0.98}. $$

Performing the division gives

$$ x \approx 0.040816. $$

To find the percentage of metal present as $$M^{3+}$$, we multiply the fraction by 100:

$$ \%\,M^{3+} = 0.040816 \times 100 \approx 4.08\%. $$

Hence, the correct answer is Option D.