The solubility of AgCl(s) with solubility product 1.6 × 10⁻¹⁰ in 0.1 M NaCl solution would be :

AgCl⇌Ag⁺+Cl⁻

Initial a 0 0
Change -S S S
Equilibrium a-S S S+0.1

ksp=1.6 × 10⁻¹⁰=[Ag⁺][Cl⁻]=S(0.1+S)

ksp value seems to be very small
S value can be ignored, with respect to 0.1 M
1.6 × 10⁻¹⁰=S × 0.1
S=1.6 × 10⁻⁹M

Hence option D is correct.