$$K_{a1}$$, $$K_{a2}$$ and $$K_{a3}$$ are the respective ionization constants for the following reactions (a), (b) and (c).
(a) $$H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-$$
(b) $$HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}$$
(c) $$H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$$
The relationship between $$K_{a1}$$, $$K_{a2}$$ and $$K_{a3}$$ is given as

The three equilibrium reactions are:

(a) $$H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-$$, with equilibrium constant $$K_{a1}$$

(b) $$HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}$$, with equilibrium constant $$K_{a2}$$

(c) $$H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$$, with equilibrium constant $$K_{a3}$$

Reaction (c) is obtained by adding reactions (a) and (b):

$$H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-$$

$$HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}$$

Adding: $$H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$$

When two equilibrium reactions are added, the equilibrium constant of the resultant reaction is the product of the individual equilibrium constants:

$$K_{a3} = K_{a1} \times K_{a2}$$

Therefore, the correct answer is Option D.