Answer
$62.51^{\circ}C$
Work Step by Step
Molality: $(0.515\ g\div 154.21\ g/mol)\div 0.015\ kg=0.223\ m$
Boiling Point Elevation: $\Delta T_b=K_b\times m=3.63^{\circ}C/m\times 0.223\ m=0.808\ ^{\circ}C$
Mixture boiling point: $61.70+0.81=62.51^{\circ}C$