Kw of water = [H+][OH-]
10^-12 = [H+][OH-] at 100 degree C
Let us assume the concentration of H+ = x
Since each water molecule dissociate into one proton and one hydroxide ion.
i.e , the quantity or the concentration of these two ions will be equal in water.
[H+] = [OH-]
10^ -12 = x.x
10^ -12 = x²
x = 10^ -6
[H+] = [OH-] = 10^ -6
pH = log [H+] = log 10^ -6 = 6 log 10 = 6
So the answer is option (a) 6.0