The solubility product constant for this salt will be 1.22 * 10^(-3).
When dissolved in water, your salt will dissociate into cations and anions according to the following equilibrium
AB_(3(s)) rightleftharpoons A_((aq))^(3+) + color(red)(3)B_((aq))^(-)
Since you're dealing with a 1.00-L solution, the molar solubility of your salt will be
C = n/V = "0.0820 moles"/"1.00 L" = "0.0820 mol/L"
Take a look at the mole ratio that exists between the species involved in the equilibrium. You have 1 mole of AB_3 dissociating to produce 1 mole of A^(3+) and color(red)(3) moles of B^(-).
You can use an ICE table to determine the value of the K_(sp).
" "AB_(3(s)) rightleftharpoons A_((aq))^(3+) + B_((aq))^(-)
I........-.............0..............0
C......-............(+x)...........(+color(red)(3)x)
E.......-.............x...............3x
By definition, K_(sp) will be
k_(sp) = [A^(3+)] * [B^(-)]^(color(red)(3)) = x * (3x)^3 = 27 * x^4
But x is actually the molar solubility of the salt, 0.0820 mol/L, which means that the value of the solubility product constant will be
K_(sp) = 27 * (0.0820)^4 = color(green)(1.22 * 10^(-3))
