For a concentrated solution of a weak electrolyte ($$K_{eq}$$ = equilibrium constant) A$$_2$$B$$_3$$ of concentration 'C', the degree of dissociation '$$\alpha$$' is

For the weak electrolyte A$$_2$$B$$_3$$ dissociating as

$$\text{A}_2\text{B}_3 \rightleftharpoons 2\text{A} + 3\text{B}$$

If the initial concentration is $$C$$ and degree of dissociation is $$\alpha$$, then at equilibrium:

$$[\text{A}_2\text{B}_3] = C(1 - \alpha) \approx C$$ (since $$\alpha$$ is small for a concentrated solution), $$[\text{A}] = 2C\alpha$$, and $$[\text{B}] = 3C\alpha$$.

The equilibrium constant is

$$K_{eq} = \frac{[\text{A}]^2[\text{B}]^3}{[\text{A}_2\text{B}_3]} = \frac{(2C\alpha)^2(3C\alpha)^3}{C}$$

$$K_{eq} = \frac{4C^2\alpha^2 \times 27C^3\alpha^3}{C} = 108C^4\alpha^5$$

Solving for $$\alpha$$,

$$\alpha^5 = \frac{K_{eq}}{108C^4}$$

$$\alpha = \left(\frac{K_{eq}}{108C^4}\right)^{1/5}$$

Hence, the correct answer is $$\alpha = \left(\dfrac{K_{eq}}{108c^4}\right)^{1/5}$$.