a)
Let's do anicebox:

              BaCrO₄              <=>            Ba²⁺ +             CrO₄^2-
I                 ___                                  0                                   0
C                                                          +x                               +x
E                                                        x                                     x

Ksp = [Ba²⁺ ][CrO₄^2-]

You're given Ksp

Ksp = x² = 2.1  10^-10
x = 1.45x10^-5M

b) This one is done the exact same way as the above, except thatthe initial concentration for chromate is not 0, but 0.004M. Thisis known as the common ion effect, and you can expect that it willdecrease the molar solubility of barium chromate.

              BaCrO₄              <=>            Ba²⁺ +             CrO₄^2-
I                 ___                                  0                                   0.004
C                                                          +x                               +x
E                                                        x                                     0.004 + x

Ksp = [Ba²⁺ ][CrO₄^2-]
2.1x10^-10 = x (0.004 + x)
You can solve the quadratic equation, but since the Ksp is sosmall, you can expect the term (0.004 + x) to essentially beunchanged. So we revise and use this equation:
2.1x10^-10 = x (0.004)
x = 5.25x10^-8