Answer
$Ag_2SO_4$ has a higher solubility.
Work Step by Step
1. Find the $AgI$ solubility:
$Kps = 8.3 \times 10^{-17} = [Ag^+]\times[I-]$
$8.3 \times 10^{-17} = x^2$
$x = 9.11 \times 10^{-9}$
2. Find the $Ag_2SO_4$ solubility:
$Kps = 1.5 \times 10^{-5}$
$Kps = [Ag^+]^2 \times [SO_4^{-2}]$
$Kps = (2x)^2 \times x$
$Kps = 4x^3$
$1.5 \times 10^{-5} = 4x^3$
$x^3 = 3.75 \times 10^{-6}$
$x = 1.55 \times 10^{-2}$
The Kps values are given in the book
