There is no "excess" I^{-}. It was present in the initial solution. *It doesn't all precipitate*. It very nearly all does, so we say it does to a first approximation, but there must always be some left in solution. Suppose the concentration of I^{-} left is x M. Then what precipitated was 0.09 - x M, and the concentration of Ag^{+} left is 0.01 + x M. So you have (0.01 + x)*x = K_{sp}. You can solve the quadratic equation for an exact value of x, but x is so much smaller than 0.01 that you can simplify it by assuming that 0.01 + x ≈ 0.01 without any loss of accuracy. (This is the "small-x approximation", and you should always check that the value you get for x *is* much smaller than 0.01(or whatever), which it clearly is here. If it isn't, you must solve the quadratic.)