#### Answer

A saturated aqueous solution of $NaBr$ at 20$^{\circ}$C contains:
$53.7g$ $NaBr$ $/ 100g$ $Solution$

#### Work Step by Step

1. The mass of the solution, in this is case, is the sum of the NaBr and the water mass.
$mass(Solution) = 116g + 100g =216g$
2. Now we know that there is 116g of NaBr for 216g of solution, and we just need to calculate this for 100 g of solution using a proportion.
$\frac{116g}{216g} = \frac{x}{100g}$
$0.537 = \frac{x}{100g}$
$x = 0.537 \times 100g = 53.7g$