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) + H ₂O (l) → O₂ (g) + I ₂ (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