(R)-2-Bromobutane is treated with aqueous NaOH under conditions that favour SN2 to give an alcohol X. X is converted to its tosylate with…

Step 1 is $\mathrm{S_N2}$ with inversion: (R)-2-bromobutane gives (S)-butan-2-ol (the priority order Br or OH $>$ ethyl $>$ methyl $>$ H is the same, so inversion flips the descriptor). Step 2 forms the tosylate at oxygen; the C–O bond is not broken, so the configuration is retained, giving the (S)-tosylate. Step 3 is another $\mathrm{S_N2}$ displacement of $\mathrm{OTs^-}$ by $\mathrm{Br^-}$ with a second inversion, giving (R)-2-bromobutane. Net: two inversions and one retention, so $\mathrm{Y}$ has the same configuration as the starting bromide.