Consider the following cell: $M\left(s\right)|M{X}_2\left(s\right),X^{-}\left(0.2M\right)|\left|M^{2+}\left(0.01M\right)\right|M\left(s\right)$ 
The EMF of the cell at 298 K is 0.236V 
[ $M{X}_2\left(s\right)$ being the sparingly soluble salt]

Pick out the correct options of the following:

$$\begin{gathered} E_{cell}=\frac{0.059}{2}\log \frac{{\left[M^{2+}\right]}_{right}}{{\left[M^{2+}\right]}_{left}} 0.236=\frac{0.059}{2}\log \frac{\left[0.01\right]}{{\left[M^{2+}\right]}_{left}} \\ {\left[M^{2+}\right]}_{left}=1\times {10}^{-10}M \therefore {\text{Solubility\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}MX}}_{2}in0.2MX=1\times {10}^{-10}M \\ {K}_{SP}ofM{X}_2=\left[M^{2+}\right]{\left[X^{-}\right]}^2=1\times {10}^{-10}\times {\left(0.2\right)}^2=4\times {10}^{-12}{M}^3 \\ {\text{Solubility\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}MX}}_{2}inwater={\left(\frac{K_{sp}}{4}\right)}^{1/3}={\left(\frac{4.0\times {10}^{-12}}{4}\right)}^{1/3}=1.0\times {10}^{-4}M \end{gathered}$$