Optical rotation of mixture
Enantiomeric excess (ee) is defined as:

$$\text{ee} = \frac{\alpha(\text{sample})}{\alpha(\text{pure enantiomer})}\tag{1}$$

Here, we are given the specific rotation of our sample ($-45^\circ$) and that of a pure enantiomer ($-53^\circ$). To calculate the ee, all you need to do is to plug these two values into equation $(1)$ and work out how to turn a fractional number into a value in $\%$.


   The answer is $85~\%~\text{ee}$.