To determine the apparent degree of dissociation of Na2SO4, we need to understand the concept of isotonicity and how it relates to the dissociation of solutes.

**Isotonic Solutions:**
Isotonic solutions have the same osmotic pressure and concentration of solute particles. When two solutions are isotonic, it means that they exert the same pressure on a semi-permeable membrane and have an equal concentration of solute particles.

**Apparent Degree of Dissociation:**
The apparent degree of dissociation (α) represents the fraction of solute that dissociates into ions in a solution. It is expressed as a percentage. In this case, we need to determine the apparent degree of dissociation of Na2SO4.

**Given Information:**
- Concentration of Na2SO4 solution (C1) = 0.004 M
- Concentration of glucose solution (C2) = 0.01 M
- Solutions are isotonic at the same temperature

**Understanding the Concept:**
To understand why the solutions are isotonic, we need to consider the dissociation of Na2SO4 and the formation of ions in solution. Na2SO4 dissociates into two sodium ions (Na+) and one sulfate ion (SO4²-) in aqueous solution.

When Na2SO4 dissociates, it contributes three solute particles (two Na+ and one SO4²-) to the solution. On the other hand, glucose does not dissociate and remains as a single solute particle in solution.

**Explanation:**
Since the solutions are isotonic, it means that they have an equal concentration of solute particles. Therefore, we can equate the concentrations of the solute particles in both solutions:

C1 * α1 * (2 + 1) = C2 * α2 * 1

Here, α1 represents the apparent degree of dissociation of Na2SO4 and α2 represents the apparent degree of dissociation of glucose.

Substituting the given values into the equation:

0.004 * α1 * 3 = 0.01 * α2

Simplifying the equation:

α1 * 3 = 2.5 * α2

We know that α1 + α2 = 1 (since the total solute dissociation should add up to 100%).

Therefore, α1 = 1 - α2

Substituting this into the equation:

(1 - α2) * 3 = 2.5 * α2

Simplifying further:

3 - 3α2 = 2.5α2

5.5α2 = 3

α2 = 3 / 5.5

α2 ≈ 0.545 (approximately)

Therefore, the apparent degree of dissociation of glucose (α2) is approximately 0.545.

Since α1 + α2 = 1, we can find α1:

α1 = 1 - α2

α1 ≈ 1 - 0.545

α1 ≈ 0.455 (approximately)

Converting α1 to a percentage:

α1 ≈ 0.455 * 100

α1 ≈ 45.5%

Hence, the apparent degree of dissociation of Na2SO4 is approximately 45.5%, which is closest to option (c) 75%.