I assume you mean #sf(10^-8color(white)(x)M)# NaOH solution.
The pH cannot be 6 because, as you correctly state, NaOH is a base or alkali. I guess you reasoned:
#sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^(-2))# at 298K
Clearly, this cannot be right as such a solution would be weakly acidic.
#sf(10^-8)#M is a fantastically dilute solution so you would expect the pH to fall towards 7 as the solution is diluted. It cannot fall below 7.
At these low concentrations you need to take into account the #sf(H^+)# formed from the auto - ionisation of water. This is done in the answer which I have given the link to.