In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ 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:
$\ce{Ag2SO4 -> 2Ag+ + SO4^{2-}}$ 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. $(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$

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.

In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ 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:
$\ce{Ag2SO4 -> 2Ag+ + SO4^2-}$ 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. $(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$

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.

In pure water, as you calculated, for the reaction

AgCl ----> Ag⁺ + Cl^- with K_sp = 1.8x10^-10

Ag₂SO₄------> 2Ag⁺ + SO₄^2- with [K_sp = 1.4x10^-5][1]

(1.34x10^-5)² x 0.25 = 4.5x10^-11 < (K_sp)_Ag2SO4

In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ 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:
$\ce{Ag2SO4 -> 2Ag+ + SO4^{2-}}$ 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. $(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$

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.