Answer
$BaBr_{2}$
Work Step by Step
Convert the mass of each constituent element into moles by dividing by its molar mass:
$2.677g\ Ba\times\frac{1mol}{137.33g}=0.01949mol\ Ba$
$3.115g\ Br\times\frac{1mol}{79.904g}=0.03898mol\ Br$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close. Then multiply by whatever factor makes them all a whole number or very close to a whole number.
$\frac{0.01949mol\ Ba}{0.01949}=1$
$\frac{0.03898mol\ Br}{0.01949}=2$
So the empirical formula for this compound is:
$BaBr_{2}$