To determine the percentage association of the solute $$\mathrm{X}$$, we need to use the boiling point elevation formula. The association of solute affects the number of particles in the solution, which in turn affects the boiling point elevation.

The boiling point elevation $$\Delta T_b$$ is given by:

$$ \Delta T_b = i \cdot K_b \cdot m $$

Where:

$$\Delta T_b$$ is the boiling point elevation
$$i$$ is the van 't Hoff factor, which indicates the number of particles into which a solute dissociates or associates in solution
$$K_b$$ is the ebullioscopic constant (boiling point elevation constant)
$$m$$ is the molality of the solution

Given data:

Boiling point of pure water = $$100^{\circ}\mathrm{C}$$
Observed boiling point of the solution = $$101.04^{\circ}\mathrm{C}$$
Boiling point elevation, $$\Delta T_b = 101.04^{\circ}\mathrm{C} - 100^{\circ}\mathrm{C} = 1.04^{\circ}\mathrm{C}$$
$$K_b$$ for water = $$0.52 \mathrm{~K} / \mathrm{m}$$
Molality, $$m = 4 \mathrm{~m}$$

We can now rearrange the boiling point elevation formula to solve for $$i$$:

$$ i = \frac{\Delta T_b}{K_b \cdot m} $$

Substitute the given values:

$$ i = \frac{1.04}{0.52 \cdot 4} $$

Calculate the value of $$i$$:

$$ i = \frac{1.04}{2.08} = 0.5 $$

The van 't Hoff factor for dimerization is given by:

$$ i = 1 - \alpha + \frac{\alpha}{2} $$

Here, $$\alpha$$ represents the degree of association. We have $$i = 0.5$$:

$$ 0.5 = 1 - \alpha + \frac{\alpha}{2} $$

Rearranging the equation to solve for $$\alpha$$:

$$ 0.5 = 1 - \frac{\alpha}{2} $$

$$ 1 - 0.5 = \frac{\alpha}{2} $$

$$ 0.5 = \frac{\alpha}{2} $$

$$ \alpha = 1 $$

Therefore, the degree of association $$\alpha$$ is 1. This indicates a 100% association of the solute $$\mathrm{X}$$ in water.

So, the correct answer is:

Option B: 100

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