To find the degree of ionization $$(\alpha)$$, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:

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

where:

$$\Delta T_b$$ is the boiling point elevation,

$$i$$ is the van't Hoff factor,

$$K_b$$ is the molal boiling point elevation constant, and

$$m$$ is the molality of the solution.

Given data:

$$\Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C}$$
Mass of $$\mathrm{AB}_2$$ = 10 g
Mass of water = 100 g = 0.1 kg
Molar mass of $$\mathrm{AB}_2$$ = 200 g/mol
$$K_b$$ = 0.52 K kg mol^-1

First, we calculate the molality (m):

$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$

$$= \frac{\frac{10}{200}}{0.1} \:\mathrm{mol\ kg}^{-1}$$

$$= 0.5 \:\mathrm{mol\ kg}^{-1}$$

Next, using the formula for boiling point elevation:

$$0.52 = i \cdot 0.52 \cdot 0.5$$

$$i = \frac{0.52}{0.52 \cdot 0.5}$$

$$i = \frac{1}{0.5}$$

$$i = 2$$

Now, the van't Hoff factor $i$ is related to the degree of ionization $$(\alpha)$$. For the electrolyte $$\mathrm{AB}_2$$, which ionizes as $$\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$$, the van't Hoff factor $i$ can be expressed as:

$$i = 1 + (n-1)\alpha$$

where

$$n$$ is the number of ions produced (which is 3 for $$\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$$)
$$\alpha$$ is the degree of ionization

Substituting $i = 2$ into the equation:

$$2 = 1 + 2 \alpha$$

$$2 - 1 = 2 \alpha$$

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

$$\alpha = 0.5$$

The degree of ionization $$(\alpha)$$ is therefore

$$0.5 \times 10^{-1} = 5 \times 10^{-2}$$.

Hence, the degree of ionization of the electrolyte $$(\alpha)$$ is approximately 5 $$\times 10^{-1}$$ (the nearest integer is 5).