Answer
$$[X^{2-}] = K_{a2}$$
This relationship exists if the ionization caused by the second proton is much smaller than the same from the first one, which happens if $K_{a2}$ is much smaller than $K_{a1}$.
Work Step by Step
When calculating the equilibrium concentrations of the solution after the first ionization we get this ICE table:
\begin{vmatrix}
& [HX^-] & [H_3O^+] & [X^{2-}] \\
Initial & x_1 & x_1 & 0 \\
Change & -x_2 & + x_2 & +x_2 \\
Equil & x_1 - x_2 & x_1 + x_2 & x_2
\end{vmatrix}
$x_1$ represents the ionization of the first proton, and $x_2$ represents the ionization of the second one.
-Writing the $K_a$ expression:
$$K_{a2} = \frac{(x_1 + x_2)(x_2)}{(x_1-x_2)}$$
If $x_1 \gt \gt x_2$:
$$K_{a2} = \frac{(x_1)(x_2)}{x_1} = x_2 = [X^{2-}]$$
