$$PbBr_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^-(aq)$$

The equilibrium constant expression for this reaction is:

$$K_{sp} = [Pb^{2+}][Br^-]^2$$

We are given that the concentrations of $Pb^{2+}$ and $Br^-$ at equilibrium are 2.5 x 10-3 M and 5.0 x 10-2 M, respectively. Substituting these values into the equilibrium constant expression, we get:

$$K_{sp} = (2.5 \times 10^{-3})(5.0 \times 10^{-2})^2 = 6.25 \times 10^{-6}$$

Therefore, the value of $K_{sp}$ for $PbBr_2$ at this temperature is 6.25 x 10-6.