Hm you have some kind of online checker for the answer? Because my answer can certainly some limitation on numerical precision. However I do think I can help as to following the correct procedure to solve the problem. First let's use a convenient notation and ignore charges and chemical structure and consider the dissociation goes like this
H2A → HA + H HA → A + H
The net equation is given by the sum of the two (the one you provided is given by the different of the two, I don't think that one is necessarily useful for the problem).
H2A → A + 2H
The eq constant for the net reaction the product of the eq constant of the partial reactions. 10-pKa1*10-pkA2 = 2.691E-10 = Knet.
Now let's solve a part of the problem that comes before you add any salt. We have to know what is the initial composition of this solution. Of course we know the analytical concentration 0.0320 M, but that doesn't tell us how it is dissociated. Because we are not interested in the concentration of the intermediate acid (HA), we can work with the net reaction instead to find the composition. Let's call C = 0.032 and pretend the acid dissociates to an extent x.
H2A → A + 2H C-x x 2x
Knet = x*(2x)2/(C-x)
This is a cubic equation, solving for x yields x = 0.000128949 (only real root).
So
[A]i = 0.000128949 [H]i = 0.000257898 [H2A]i = 0.031871051
(i means initial, prior to adding the salt, I solved it using Wolfram Alpha. If you want to do it by hand, consider [H2A] seems unchanged because x is a very small number to subtract and solve the equation by taking the cubic root after considering (C-x) ≈ (C))
When you perturb this system by adding salt, you have a new non-equilibrium composition which is:
[A]i = 0.000128949 +M [H]i = 0.000257898 [H2A]i = 0.031871051
Where M denotes the concentration increase by adding salt to the solution. This system is perturbed and needs now to find a new state of equilibrium. By Chatelier's rule I know equilibrium will shift in favor of the left side (H2A → A + 2H), so [H] has to go down. The new composition will be
[A]f = 0.000128949 +M - y [H]f = 0.000257898 - 2y [H2A]f = 0.031871051 + y
(I'm using y to distinguish using x before, the "2y" accounts for the stoichiometry of the reaction).
But I do know the final concentration [H]f = 10-pH = 1.422E-006
So
1.422E-006 = 0.000257898 - 2y
y = 0.000128238
So
[A]f = 0.000000711 +M [H]f = 0.000001422 [H2A]f = 0.031999289
Again
Knet = 2.691E-10 = (0.000000711 +M )* 0.0000014222 / 0.031999289 M = 4.2584814638
Moles = 4.2584814638 * 670.010-3 = 2.8531825807 moles Mass = 2.8531825807 248.32 = 708.50 grams
There may be some numerical mistake I did but I am pretty certain you're guaranteed with this procedure. I hope this all makes sense to you.