The molar solubility of calcium chromate in a 0.170 M calcium acetate solution is M.?

Common ion effect using calcium chromate and calcium acetate....

With a common ion present, in this case the calcium ion, the solubility of CaCrO4 will be less. This can be "explained" using Le Chatelier's principle.

CaCrO4(s) <==> Ca2+ + CrO4^2- ................. Ksp = ???

If additional calcium ion is added (in the form of Ca(C2H3O2)2), then the equilibrium will move to the left as the concentration of Ca2+ moves closer to what it was originally. This will reduce the solubility of CaCrO4.

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The concentration of CaCrO4 at 20C is 22.5g / 1L.... or .... 22.5g / 156.072 g/mol / 1L= 0.144M
See solubility data here: https://en.wikipedia.org/wiki/Calcium_chromate
Ksp = [Ca2+][CrO4^2-] = 0.144^2 = 0.0208

CaCrO4(s) <==> Ca2+ + CrO4^2- ................. Ksp = 0.0208
......................... 0.314M...0.144M .................. initial
......................... -x ........... -x .......................... change
........................ 0.314-x....0.144-x .................. equilibrium

Ksp = [Ca2+][CrO4^2-]
0.0208 = (0.314-x)(0.144-x)
x = 0.0616
[Ca2+] = 0.252M
[CrO4^2-] = 0.0824M

The molar solubility of CaCrO4 will be 0.0824M

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Note: Other sources give the Ksp as 7.1x10^-4. Simply substitute 7.1x10^-4 in place of the Ksp derived from Wikipedia and compute the new molar solubility of CaCrO4.

CaCrO4(s) <==> Ca2+ + CrO4^2- ................. Ksp = 7.1x10^-4
......................... 0.314M...0.144M .................. initial
......................... -x ........... -x .......................... change
........................ 0.314-x....0.144-x .................. equilibrium

Ksp = [Ca2+][CrO4^2-]
7.1x10^-4 = (0.314-x)(0.144-x)
x =0.140
[Ca2+] = 0.174M
[CrO4^2-] = 0.0041M

The molar solubility of CaCrO4 will be 0.0041M

Refer to: https://en.wikipedia.org/wiki/Calcium_acetate
The solubility of calcium acetate at 20 °C is 34.7 g/100 mL water.

It is quite soluble.
Solubility equilibria (Ksp) are only applied to sparingly soluble salts. Therefore, it is impossible to use solubility equilibrium to calculate the molar solubility as Ksp is undefined for a quite soluble salt.