What mass of `"NaBr"` is needed to prepare `"50.00 mL"` of a `6.5%` (m/v) solution in distilled water?

A solution's mass by volume percent concentration, `"m/v %"`, tells you the number of grams of solute present in exactly `"100.0 mL"` of the solution.

So all you need to know in order to figure out a solution's mass by volume percent concentration is the number of grams of solute present for every `"100..0 mL"` of the solution.

In your case, you know that you're dealing with a `6.5%` `"m/v"` solution, which means that every `"100.0 mL"` of this solution contain `"6.5 g"` of sodium bromide, the solute.

So you can say that in order to get the equivalent of `"6.5 g"` of sodium bromide in `"100.0 mL"` of this solution, you need `"50.00 mL"` of the solution to contain

`50.00 color(red)(cancel(color(black)("mL solution"))) * "6.5 g NaBr"/(100.0color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("3.3 g NaBr")))`

The answer is rounded to two sig figs, the number of sig figs you have for the mass by volume percent concentration.

To make this solution, you would dissolve `"3.3 g"` of sodium bromide in `"20.00 mL"` of distilled water, then add enough water to ensure that the final volume of the solution is equal ot `"50.00 mL"`.