SN1, SN2 and enantiomeric excess
In this case, the absolute configuration remains same (as R) for $\mathrm{S_N1}$ and changes (into S) for $\mathrm{S_N2}.$ This is because the priority order of non substituted groups (wrt $\ce{C2})$ remains the same before and after substitution of $\ce{Br}$ by $\ce{OH}.$

Let the initial mole fraction of (R)-2-bromobutane be 1. After the reaction, let the mole fractions of R- and S-2-bromobutane be $x$ and $1-x$, respectively.

According to question condition, as $\mathrm{S_N1}$ proceeds twice as fast as $\mathrm{S_N2},$ we get

$$x = 2(1 - x) \quad\implies\quad x = 2/3$$

So we have excess of R-enantiomer, and the specific rotation is

$$\frac{2}{3}\cdot (+13.5) + \frac{1}{3}\cdot (-13.5) = +4.5$$