What is the maximum number of silver(Ⅰ) ions that can be present dissolved in one litre of a $\pu{0.003 M}$ $\ce{Na2S}$ solution?
According to my book, silver(I) reacts with sulfide producing $\ce{Ag2S}$ with the solubility product constant
$$K_\mathrm{sp}(\ce{Ag2S}) = [\ce{Ag+}]^2\,[\ce{S^2-}] = \pu{8E-51}.$$
Now here is what I don't understand. I have seen several questions where they ask to solve for the maximum solubility of a substance in water, in this case $\ce{Ag2S}$. The usual thing they do is set up an equation with $x$, like $K_\mathrm{sp} = (2x)^2\,x.$
But in our case, we are given an initial concentration $c_0(\ce{Na2S}) = \pu{0.003 M}$, from which we know that the concentration $c_0(\ce{S}) = \pu{0.003 M}$. We also know the $K_\mathrm{sp}$ value, and they ask specifically for the silver(I) ions, not silver(I) sulfide.
So I am not sure if I need to proceed like above, i.e. set up an equation with $x$ and solve for it, because it seems to me that would give the maximum solubility of $\ce{Ag2S}$, not the maximum number of silver ions.