Ksp calculation of Cr(OH)2
Your calculation of $[\ce{OH-}] = 10^{-5.51}\ \mathrm{mol\ l^{-1}}$ is correct.

The pH is well above 7. Therefore, the present $[\ce{OH-}]$ is mainly a result of the dissociation of $\ce{Cr(OH)2}$:

$$\ce{Cr(OH)2 &lt;=&gt; Cr^2+ + 2 OH- }$$

Thus, you know 
$[\ce{Cr^2+}] \approx \tfrac{1}{2} [\ce{OH-}] = \tfrac{1}{2}\cdot10^{-5.51}\ \mathrm{mol\ l^{-1}} $.

$K_\text{sp} = [\ce{Cr^2+}][\ce{OH-}]^2 = \tfrac{1}{2}\cdot10^{-5.51} \cdot \left(10^{-5.51}\right)^2 = 1.48\cdot 10^{-17}$