Given that resistance is 8.772 MΩ,
R = l/(kA) where l is the length and A is the area of the cell. k is the conductivity.
I get k= 2.85 x 10-8 /Ωcm
So molar conductivity is Λ=k/c where c is the concentration. Here assuming 1g/cc as the density of water, I get c = (1/18) mol/cc.
This gives Λ = 51.3 x 10-8 cm2/Ωmol.
knowing that, I can find the degree of dissociation as α= Λ/Λ∞
Λ∞ = 415 cm2/Ωmol.
So α = 1.23 x 10-9.
Thus the water product is (c*α)2 which is 4.67 x 10-15 mol2/L2.