The solubility product of a salt having general formula $$MX_2$$, in water is $$4 \times 10^{-12}$$. The concentration of M$$^{2+}$$ ions in the aqueous solution of the salt is

The salt dissociates in water as $$MX_2 \rightarrow M^{2+} + 2\,X^-$$.

If the molar solubility of the salt is $$s$$ (mol L$$^{-1}$$), the equilibrium concentrations become
$$[M^{2+}] = s$$ and $$[X^-] = 2s$$.

For a salt of the type $$MX_2$$ the solubility-product expression is
$$K_{sp} = [M^{2+}]\,[X^-]^2$$.

Substituting the equilibrium concentrations:
$$K_{sp} = s \,(2s)^2 = 4\,s^3$$.

Given $$K_{sp} = 4 \times 10^{-12}$$,
$$4\,s^3 = 4 \times 10^{-12} \;\;\Longrightarrow\;\; s^3 = 10^{-12}$$.

Taking the cube root:
$$s = 10^{-4}\ \text{mol L}^{-1}$$.

Therefore, the concentration of $$M^{2+}$$ ions in the saturated solution is $$[M^{2+}] = s = 1.0 \times 10^{-4}\ \text{M}$$.
Option B which is: $$1.0 \times 10^{-4}\ \text{M}$$.