At normal body temperature, 37 degrees Celsius, the value of Kw for water is 2.5 x 10^-14. Calculate [H3O+] and [OH-] at this temperature. Is pure water at 37 degrees Celsius acidic, neutral or basic?

I was told that this was a complicated ques. Since they only give you Kw (2.5 x 10^-14)
But the answer in my book says...
[H3O+] = [OH-]= 1.6 x 10^-7

It's not a complicated question.

The product of [H3O+] and [OH-] will always equal the Kw at the temperature specified. Most often, we use 1 x 10^-14 as the Kw. That's probably at room temperature.

This is the equilibrium:

2H_2O \rightleftharpoons H_3O^+ + OH^-

I'm sure you see that for every hydrogen ion (hydronium ion) that is produced a hydroxide ion is also produced. Hence, it's neutral. Also, the concentrations of the two species will be the same. Therefore water is "neutral" -- neither acidic nor basic.

[H_3O^+][OH^-]=[OH-]^2=[H_3O^+]^2 \approx 2.5 \times 10^-14\,at\,37\,C

The concentration of either ion is

[H_3O^+] = [OH^-] = \sqrt {2.5} \times 10^{-7} \approx 1.58 \times 10^{-7}