We start with [Ag+] = 1.00 mol L^{-1} and [Cl^{-}] = 0.250 mol L^{-1}
The K_{sp} equilibrium is
AgCl(s) <---> Ag^{+}(aq) + Cl^{-}(aq)
and it lies far to the left given the small value of K_{sp} = [Ag^{+}][Cl^{-}] = 1.76 x 10^{-10}
So, at equilibrium, we will have [Ag+] = 1.00 - x mol L^{-1} and [Cl^{-}] = 0.250 - x mol L^{-1}
and for the equation (1.00 - x)(0.250 - x) = 1.76 x 10^{-10} to be solvable, noting x is the extent to which the system moved left (and hence x < 0.250), we will have [Cl^{-}] = being very close to 0, x very close to 0.250, and so [Ag^{+}] very close to 0.750 mol L^{-1}.
At equilibrium, [Ag^{+}] = 0.750 mol L^{-1} (3 sig. fig.)
[Cl^{-}] = K_{sp} / [Ag^{+}] = 1.76 x 10^{-10} / 0.750 = 2.34 x 10^{-10} mol L^{-1} (3 sig. fig.)
Check:
x = 0.250 - [[Cl^{-}] = 0.249 999 999 76... mol L^{-1}
and so actually [Ag^{+}] = 0.750 000 000 23... mol L^{-1}, which is consistent with the approximation above that [Ag^{+}] = 0.750 mol L^{-1} (3 sig. fig.)
which fits above results.
Edited to emphasise result for concentration of silver(I) ions.