Calculate the concentration of I− in a solution obtained by shaking 0.10 M KI with an excess of AgCl(s)?

This involves two solubility equilibria (AgI and AgCl).  These two equilibria are then combined to give the following equilibrium:
I⁻(aq) + AgCl(s) ⇌ Cl⁻(aq) + AgI(aq)
Note that the equation must be written in complete ionic form.
You may then assume the equilibrium goes almost completely to the right due to the very great equilibrium constant.

My answer is as follows:

Refer to: https://www.chm.uri.edu/weuler/chm112/refmater/KspTable.html
Ksp(AgCl) = 1.8 × 10⁻¹⁰
Ksp(agI) = 8.5 × 10⁻¹⁷
(Ksp values from different sources may be slightly different.)

(1): AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)          K₁ = Ksp(AgCl)
(2): Ag⁺(aq) + I⁻(aq) ⇌ AgI(s)               K₂ = 1/Ksp(AgI)

(1) + (2), and cancel Ag⁺(aq) on each side:
I⁻(aq) + AgCl(s) ⇌ Cl⁻(aq) + AgI(aq)     Kc
Kc = K₁K₂ = Ksp(AgCl)/Ksp(AgI) = (1.8 × 10⁻¹⁰)/(8.5 × 10⁻¹⁷) = 2.12 × 10⁶

Since Kc is very large, the reaction goes almost completely to the right.  In other words, I⁻ ions almost completely reacts with AgCl solid to from Cl⁻ ions and AgI(s).

Hence, [Cl⁻] at equilibrium ≈ [I⁻]ₒ = 0.10 M

At equilibrium: Kc = [Cl⁻]/[I⁻]
0.10/[I⁻] = 2.12 × 10⁶
[I⁻] = 0.1/(2.12 × 10⁶) M
[I⁻] at eqm = 4.7 × 10⁻⁸ M