Predict the order of ionic radius
You're right about the expected order. From largest to smallest the expected order would be:

$\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe]

$\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_{35}Br^-}\text{  }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_8O^{2-}}\text{   }\text{ }$ = [He] $2s^2\text{ }2p^6$ = [Ne]

$\ce{_9F^-}\quad$ = [He] $2s^2\text{ }2p^6$ = [Ne]

You expressed the idea poorly though. $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ have the same number of electrons. However $\ce{_9F^-}$ has an extra proton which should draw the electrons in closer. 

There are a lot of ways to measure atomic/ionic size. No idea why the book would have $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ flipped.