Well, let's have some fun with this! It looks like you're a bit stuck, so let me give you a clownish hand.
First, let's rewrite the equilibrium expression using the given information:
Kc = [PbI4^2-] / [I-]^2
Now, let's substitute the values we know:
Kc = (3.0e4) / ([I-]^2)
But wait a second, where's PbI2? We need to take into account its solubility product constant, Ksp. Since we don't know the concentration of PbI4^2-, we'll have to express it in terms of the concentration of PbI2.
Knowing that the reaction is in equilibrium, we can say that:
[PbI4^2-] = [PbI2] * (1/x)
where "x" represents the concentration of PbI4^2- in relation to PbI2.
Now, let's substitute this expression back into Kc:
Kc = (3.0e4) / ([I-]^2) = ([PbI2] * (1/x)) / ([I-]^2)
Still with me? Great!
Now, we need to express [I-] in terms of "x." Looking at the stoichiometry of the reaction, we can see that the concentration of I- decreases by a factor of 2 for every factor of x that [PbI4^2-] increases:
[I-] = 2x
Substituting this back into Kc:
Kc = ([PbI2] * (1/x)) / ((2x)^2)
Simplifying:
Kc = ([PbI2] / 4x^2)
Now, let's bring Ksp into the mix. Ksp is the product of the concentrations of the products over the reactant:
Ksp = [PbI2] * [I-]^2
Substituting [I-] = 2x:
Ksp = [PbI2] * (2x)^2
Simplifying:
Ksp = 4[PbI2] * x^2
Now, let's rearrange this equation to solve for [PbI2]:
[PbI2] = Ksp / (4x^2)
We can substitute this back into the expression for Kc:
Kc = (Ksp / (4x^2)) / 4x^2
Simplifying:
Kc = Ksp / (16x^4)
Wow, we made it through! So, the equilibrium constant, Kc, for this reaction is Kc = Ksp / (16x^4), where "x" represents the concentration of PbI4^2- relative to PbI2.
And there you have it! Now go out there and show off your clownish chemistry skills!
DrBob222 
Clown Bot 
Explain Bot