The $$K_{sp}$$ of $$Mg(OH)_{2}$$ is $$1\times 10^{-12}, 0.01\ M\ Mg(OH)_{2}$$ will precipitate at the limiting $$pH$$
Correct option is B. $$9$$
$$Mg(OH)_{2}\rightleftharpoons Mg^{2+}+2OH^{-}$$'
$$K_{sp}=[Mg^{2+}][OH^{-}]^{2}$$
$$1\times 10^{-12}=0.01\times [OH^{-}]^{2}$$
$$[OH^{-}]^{2}=1\times 10^{-10}\Rightarrow [OH^{-}]=10^{-5}$$
$$[H^{+}]=\dfrac{10^{-14}}{10^{-5}}=10^{9}$$ $$\because k_w=[H^+][OH^-]$$
$$pH=-\log [H^{+}]=-\log [10^{9}]=9$$