Finding the initial concentration
This is a titration problem for quantitatively determining the concentration of a solution. 

What reaction does occur?


   $\ce{AgNO3 (aq) + NaBr (aq) &lt;=&gt; AgBr v + Na+ (aq) + NO3- (aq)}$
 or essentially $\ce{Ag+ + Br- &lt;=&gt; AgBr v}$  


Why is the solubility product important? 


   The solubility product tells you about the extent of the reaction. In this particular case it tells you, that you reached equilibrium between ions in solution and the precipitated salt. It tells you exactly the product of concentrations in a saturated solution.


What can you tell about the equilibrium state at the point when the first precipitate falls?


   The solubility product is matched, hence $\ce{[Ag^+][Br^-]}&lt;K_s=7.7\cdot10^{-13}~\mathrm{\left(\frac{mol}{L}\right)}^2$


What is the amount of bromine ions added to the solution?


   $n(\ce{Br^-}) = V(\ce{NaBr})\cdot c(\ce{NaBr}) = 0.020~\mathrm{mL}\cdot 0.001~\mathrm{\frac{mol}{L}} = 2\cdot10^{-5}~\mathrm{mol}$  


What can you tell about the concentrations in the final mixture?
First, what is the concentration of bromide ions in this mixture?


   $V_0(\ce{AgNO3}) = 0.5~\mathrm{L}$, $V(\ce{NaBr}) = 0.02~\mathrm{L}$, $V_t = 0.52~\mathrm{L}$
 $c_t(\ce{Br^-}) = \frac{n(\ce{Br^-}}{V_t} \approx 3.8\cdot10^{-5}~\mathrm{\frac{mol}{L}}$


Second, what can you tell about the concentration of silver ions in the final mixture?


   $c(\ce{Ag^+}) = \frac{K_s}{c(\ce{Br^-})} = \frac{K_s}{\frac{V(\ce{NaBr})}{V_t}\cdot c(\ce{NaBr})} = \frac{K_s\cdot V_t}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 2\cdot10^{-7}~\mathrm{\frac{mol}{L}}$ 


What is the number of moles of silver ions in the final mixture?


   $n(\ce{Ag^+}) = c(\ce{Ag^+})\cdot V_t = \frac{K_s\cdot V_t^2}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 1\cdot10^{-7}~\mathrm{mol}$


What is the initial concentration of the silver nitrate solution?


   $c_0(\ce{AgNO3}) = \frac{n(\ce{Ag^+})}{V_0(\ce{AgNO3})} = \frac{K_s\cdot V_t^2}{V_0(\ce{AgNO3})\cdot V(\ce{NaBr})\cdot c(\ce{NaBr})} = 2.08\cdot10^{-8}~\mathrm{\frac{mol}{L}}$