How many moles of sodium iodide are needed to remove 5.95×10−6 mol of o3?

You have to find the reaction involving sodium iodide and ozone or O₃. If you search in the internet, the reaction would be

O₃(g) + 2 NaI(aq) + H₂O(l) → O₂(g) + I₂(g) + 2 NaOH(aq) 

Using the stoichiometric coefficients in the reaction, the solution would be

5.95×10⁻⁶ mol O₃ * (2 mol NaI/1 mol O₃) = 1.19×10⁻⁵ mol NaI