MX > M$$_3$$X > MX$$_2$$

$$\Rightarrow~~\mathrm{MX} \to$$ $$\mathrm{\mathop {{M^ + }}\limits_S + \mathop {{M^ - }}\limits_S}$$

Solubiliity, (S$$_1$$) of

$$MX = \sqrt {{K_{sp}}} = \sqrt {4 \times {{10}^{ - 8}}} $$

$$ = 2 \times {10^{ - 4}}\,M$$

$$ \Rightarrow M{X_2} \to \mathop {{M^ + }}\limits_S + \mathop {2{X^ - }}\limits_S $$

$${K_{sp}} = {(s)^1}{(2s)^2} = 4{s^3}$$

$$S = {\left( {{{3.2 \times {{10}^{ - 14}}} \over 4}} \right)^{{1 \over 3}}}$$

$${K_{sp}} = 2 \times {10^{ - 5}}\,M$$

$$ \Rightarrow {M_3}X \to \mathop {3{M^ + }}\limits_{3S} + \mathop {{X^{3 - }}}\limits_S $$

$${K_{sp}} = {(3s)^3}{s^1} = 27{s^4}$$

$$S = {\left( {{{2.7 \times {{10}^{ - 15}}} \over {27}}} \right)^{{1 \over 9}}}$$

$$ = {({10^{ - 16}})^{{1 \over 2}}}$$

$$S = {10^{ - 4}}\,M = 1 \times {10^{ - 14}}\,M$$

Hence, option (D) is correct.

MX > M$$_3$$X > MX$$_2$$