This is the Finkelstein reaction.
Sodium iodide (NaI) reacts with methyl chloride ($\mathit{C}{\mathit{H}}_3\mathit{C}\mathit{l}$).
Iodide ion $\left(\mathit{I}-\right)$ replaces chloride $\left(\mathit{C}\mathit{l}-\right)$.
Dry acetone is used because NaCl is insoluble in acetone, so it precipitates out, driving the reaction forward.
Products:
A = $\mathit{C}{\mathit{H}}_3\mathit{I}$ (methyl iodide)
B = NaCl (precipitate)