Find the h3o+ and the oh- from the following values

Missing question on internet:
a) pH = 5.
pH = -log[H₃O⁺].
[H₃O⁺] = 10∧(-pH).
[H₃O⁺] = 10⁻⁵ M.
The Kw (the ionic product for water) at 25°C is 1·10⁻¹⁴ mol²/dm⁶ or 10⁻¹⁴ M².
Kw = [H₃O⁺] · [OH⁻].
[OH⁻] = Kw ÷ [H₃O⁺].
[OH⁻] = 10⁻¹⁴ M² ÷ 10⁻⁵ M = 10⁻⁹ M.
b) pH = 1.82.
[H₃O⁺] = 10∧(-1.82) = 1.5·10⁻² M.
[OH⁻] = 10⁻¹⁴ M² ÷ 1.5·10⁻² M = 6.6·10⁻¹³ M.
c) pH = 10.65.
[H₃O⁺] = 10∧(-10.65) = 2.23·10⁻¹¹ M.
[OH⁻] = 10⁻¹⁴ M² ÷ 2.23·10⁻¹¹ M = 4.46·10⁻⁴ M.

The answer options is d