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} cm^{2}/Ωmol.

knowing that, I can find the degree of dissociation as α= Λ/Λ^{∞}

Λ^{∞} = 415 cm^{2}/Ωmol.

So α = 1.23 x 10^{-9}.

Thus the water product is (c*α)^{2} which is 4.67 x 10^{-15} mol^{2}/L^{2}.