For the reaction RCH 2 Br + I - ⟶  Acetone  RCH 2 I  major  + Br - The correct…

$\mathrm{R}-{\mathrm{CH}}_2-\mathrm{Br}+{\mathrm{I}}^{-}\stackrel{\text{ Acetone }}{⟶}\mathrm{R}-{\mathrm{CH}}_2-\mathrm{I}+{\mathrm{Br}}^{-}$

As mentioned the reaction follows ${\mathrm{S}}_{\mathrm{N}}2$  reaction mechanism. The Finkelstein reaction classically involves the conversion of Alkyl bromides into Alkyl iodides by the treatment with a solution of Sodium iodide in Acetone. $\mathrm{NaBr}$ is not readily soluble in acetone and hence reaction shifts in forward reaction. For ${\mathrm{S}}_{\mathrm{N}}2$ reactions, transition state formed is less polar than the localised anion.