The first thing that you need to do here is to figure out the molar solubility of silver bromide in water, i.e. the number of moles of silver bromide that can dissolve in
So, you know that silver bromide is considered insoluble in water, so you can say that when this salt is dissolved in water, an equilibrium is established between the undissolved solid and the solvated ions.
#"AgBr"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Br"_ ((aq))^(-)#
By definition, the solubility product constant,
#K_(sp) = ["Ag"^(+)] * ["Br"^(-)]#
Now, if you take
#K_(sp) = s * s^2#
which, in your case, is equal to
#5.0 * 10^(-7) = s^2#
#s = sqrt(5.0 * 10^(-7)) = 7.07 * 10^(-4)#
This tells you that you can only hope to dissolve
Consequently, you can say that
#100 color(red)(cancel(color(black)("L solution"))) * (7.07 * 10^(-4)color(white)(.)"moles AgBr")/(1color(red)(cancel(color(black)("L solution")))) = color(darkgreen)(ul(color(black)(7 * 10^(-2)color(white)(.)"moles AgBr")))#
The answer is rounded to one significant figure, the number of sig figs you have for the volume of water.