Solubility of Ag3AsO4 in 0.02 M K3AsO4
You start of by calculating solubility with one approximation: that all $\ce{AsO4^3-}$ ions are obtained by dissolving the potassium salt:
Let $S$ be the molar solubility:
$$K_\mathrm{sp} = 0.02 \times 27 S^3$$
$$S = \pu{0.0223 M}$$
Then, you calculate the solubility by taking in account the concentration of arsenate ion obtained by dissolution of silver arsenate:
$$K_\mathrm{sp} = 27 S^3 \times (S+0.02) = 27 S^4 + 0.54S^3$$
Solving this equation by applying the iteration method gives the value of $S = \pu{0.018M}$.
So the relative error is $23.9~\%$.