1. Write the dissolution equilibrium

For chromium(III) hydroxide:

$Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq)$

2. Let the solubility be $s$ mol L$^{-1}$

Then at saturation:

$[Cr^{3+}] = s, \qquad [OH^-] = 3s$

3. Write the solubility product expression

$K_{sp} = [Cr^{3+}][OH^-]^3$

Given:

$K_{sp} = 6.0 \times 10^{-31}$

So,

$6.0 \times 10^{-31} = s(3s)^3$

$6.0 \times 10^{-31} = 27s^4$

$s^4 = \frac{6.0 \times 10^{-31}}{27}$

4. Find $[OH^-]$ directly

Since $[OH^-] = 3s$, let $x = [OH^-]$. Then:

$[Cr^{3+}] = \frac{x}{3}$

Using $K_{sp}$:

$K_{sp} = \frac{x}{3} \cdot x^3 = \frac{x^4}{3}$

Hence,

$\frac{x^4}{3} = 6.0 \times 10^{-31}$

$x^4 = 18 \times 10^{-31}$

Therefore,

$[OH^-] = (18 \times 10^{-31})^{1/4}$

5. Match with the options

This corresponds to Option C.

Final Answer:

$\boxed{(18 \times 10^{-31})^{1/4}}$