The bromide shown is treated first with $\mathrm{NaI}$ in dry acetone (SN2), and the iodide so formed is then treated with $\mathrm{NaCN}$ in DMSO (SN2). Which structure is the final product?

Each SN2 step inverts the stereocentre. Two inversions return the original spatial arrangement (net retention): (R)-2-bromobutane gives (S)-2-iodobutane, which gives (R)-2-methylbutanenitrile (CN attacks through carbon, so a nitrile, not an isocyanide). The enantiomeric nitrile would need an odd number of inversions, and the iodide is only the intermediate.