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))#
