1. Write the dissolution equilibrium

For the salt $A_3B_2$:

$A_3B_2(s) \rightleftharpoons 3A^{2+} + 2B^{3-}$

Let the molar solubility be $s$ mol L$^{-1}$.

Then at equilibrium:

$[A^{2+}] = 3s, \qquad [B^{3-}] = 2s$

2. Write the solubility product expression

$K_{sp} = [A^{2+}]^3 [B^{3-}]^2$

Substituting the concentrations:

$K_{sp} = (3s)^3(2s)^2$

$K_{sp} = 27s^3 \cdot 4s^2 = 108s^5$

3. Relate molar solubility to given solubility in g L$^{-1}$

Given solubility is $x$ g L$^{-1}$ and molar mass is $M$ g mol$^{-1}$, so

$s = \frac{x}{M}$

Therefore,

$K_{sp} = 108\left(\frac{x}{M}\right)^5$

Comparing with

$K_{sp} = a\left(\frac{x}{M}\right)^5$

we get

$a = 108$

4. Comparison with stored answer

Stored correct answer = $108$

This matches the derived answer.