Answer
The molar solubility is $6\times10^{-4} M$
Work Step by Step
$Ksp=6.0\times10^{-16}$
When a solution is buffered, its pH hardly changes. For simplicity, let's assume that pH in this solution does not change at all when an amount of $Ni(OH)_{2}$ is added. Because pH doesn't change, pOH doesn't change, too. Since, we are able to calculate $[OH^{-}]$.
$pH + pOH=14$
$pOH=14-pH=14-8.0=6.0$
$-> [OH^{-}]=10^{-pOH}=10^{-6} M$
Note again that this solution is buffered; so no matter how much $Ni(OH)_{2}$ is added, [$OH^{-}$] would still stay constant (see section 17.2 for more details about buffers).
Now, suppose $x$ M is the amount of $Ni(OH)_{2}$ solid needed to dissolve and form a saturated solution. We can construct the following table:
$Ni(OH)_{2} (s) Ni^{2+} (aq) + 2OH^{-} (aq)$
Initial: $[Ni(OH)_{2}]=x M, [Ni^{2+}]= 0 M, [OH^{-}]=10^{-6} M$
Equilibrium: $[Ni(OH)_{2}]=0 M, [Ni^{2+}]= x M, [OH^{-}]=10^{-6} M$
As a saturated solution, $Ksp=[Ni^{2+}][OH^{-}]^2=6.0\times10^{-16}$
$-> x\times(10^{-6})^2=6.0\times10^{-16}$
$-> x=6.0\times10^{-4} M$ (use a calculator)
In conclusion, the molar solubility for nickel (II) hydroxide in this scenario is $6\times10^{-4} M$.
