(R)-2-Bromobutane reacts with $\mathrm{NaI}$ in dry acetone to give $\mathrm{X}$, and $\mathrm{X}$ reacts with sodium ethanoate to give $\mathrm{Y}$. Assume each step is a single clean SN2 reaction with inversion. The product $\mathrm{Y}$ is

Step 1 gives (S)-2-iodobutane, since the priorities $\mathrm{I}>\mathrm{C}{}_2\mathrm{H}{}_5>\mathrm{CH}{}_3>\mathrm{H}$ keep the same ranking pattern as for Br and inversion changes R to S. In step 2 the oxygen of the ethanoate group again has the highest priority, followed by $\mathrm{C}{}_2\mathrm{H}{}_5>\mathrm{CH}{}_3>\mathrm{H}$, and a second inversion converts S back to R. Two inversions give net retention of the spatial arrangement, so $\mathrm{Y}$ is (R)-butan-2-yl ethanoate.