How to calculate the solubility of AgCl in a MgSO₄ solution?
In pure water, as you calculated, for the reaction
$\ce{AgCl -&gt; Ag+ + Cl-}$ with $K_\mathrm{sp} = 1.8\cdot10^{-10}$
 the concentration of $\ce{Ag+}$, i.e. $\ce{[Ag+]}$ will be $1.34\cdot10^{-5}~\mathrm{M}$.

In a solution of $0.25~\mathrm{M}$ $\ce{MgSO4}$ (which dissociates almost completely), there's an additional equilibrium here:
$\ce{Ag2SO4 -&gt; 2Ag+ + SO4^2-}$ with $K_\mathrm{sp} = 1.4\cdot10^{-5}$  

Looking at the $K_\mathrm{sp}$ of both the reactions, we can see that $\ce{AgCl}$, having the lesser solubility, decides the $\ce{[Ag+]}$. Here we can even check if the concentration of $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$ and $\ce{[SO4^{2-}]} = 0.25~\mathrm{M}$ exceeds the solubility limit of $\ce{Ag2SO4}$. It doesn't. 
$(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} &lt; (K_\mathrm{sp})_{\ce{Ag2SO4}}$

The final composition of the solution will be $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Cl- ]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Mg^2+]} = 0.25~\mathrm{M}$ and $\ce{[SO4^2- ]} = 0.25~\mathrm{M}$.

In conclusion, the $\ce{[Ag+]}$ is the same for both the cases.