Calcium oxalate, cac2o4 (m = 128.1), dissolves to the extent of 0.67 mg l–1 . what is its ksp?

First balance the equation:
CaC2O4 (s) -> Ca2+ (aq) + C2O4 2- (aq)

from the balanced equation, write Ksp:
Ksp=[Ca2+][C2O4]  (Calcium oxalate is not included because it is a solid)

convert mg to g and then g/L into Molarity (Mol/L) to find the concentrations used in finding Ksp:
0.67mg/L x (1 g/ 1000 mg) x (1 mol/ 128.1 g) = 5.23e-6 M

Set up an ICE table from the balanced equation with previous calculation as the amount that has dissociated: (again, you can ignore Calcium oxalate because it is a solid and not included in Ksp)

    CaC2O4 (s) -> Ca2+ (aq) + C2O4 (aq)
I    -                      0                    0
C  -                       +5.23e-6       +5.23e-6
E  -                       +5.23e-6       +5.23e-6

You know have the concentrations of the species necessary for Ksp:
Ksp=(Ca2+)(C2O4)
=(5.23e-6)(5.23e-6)

Ksp = 2.74e-11