The solubility of $$Ca(OH)_2$$ in water is:
[Given: The solubility product of $$Ca(OH)_2$$ in water $$= 5.5 \times 10^{-6}$$]

For $$Ca(OH)_2$$, the dissociation in water is: $$Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-$$. If the solubility is $$s$$, then $$[Ca^{2+}] = s$$ and $$[OH^-] = 2s$$.

The solubility product is $$K_{sp} = [Ca^{2+}][OH^-]^2 = s \cdot (2s)^2 = 4s^3$$.

Given $$K_{sp} = 5.5 \times 10^{-6}$$, we have $$4s^3 = 5.5 \times 10^{-6}$$, so $$s^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6}$$.

Taking the cube root: $$s = (1.375 \times 10^{-6})^{1/3} = (1.375)^{1/3} \times 10^{-2}$$. Now $$(1.375)^{1/3} \approx 1.112$$.

So $$s \approx 1.11 \times 10^{-2}$$ mol/L.

Therefore, the solubility of $$Ca(OH)_2$$ is $$1.11 \times 10^{-2}$$, which is option (3).