So, what's shown here is the ion product of pure water: that is, the product of the concentrations of hydronium and hydroxide ions in pure water at 25 °C. By this relation, if you know the [H₃O⁺], you can calculate the [OH⁻], and vice-versa.
Since [H₃O⁺] × [OH⁻] = 1.0 × 10⁻¹⁴, [OH⁻] = (1.0 × 10⁻¹⁴)/[H₃O⁺].
Substituting the given [H₃O⁺] as 1.25 × 10⁻² M:
[OH⁻] = (1.0 × 10⁻¹⁴)/(1.25 × 10⁻² M) = 8.0 × 10⁻¹³ M.
