The number of Faradays required to completely deposit magnesium from 1 L of $0.1 \mathrm{M} {\mathrm{MgCl}}_2\text{ aq. }$ solution is 

Number of moles of MgCl₂ in 0.1 M (in 1 L) is 0.1 moles.

The equivalent weight of Mg is:

$$\begin{gathered} {\mathrm{Mg}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mg} \\ \mathrm{Equivalent} \mathrm{weight}=\frac{24}{2}=12 \end{gathered}$$

According to Faraday's law of electrolysis, one mole of Mg requires 2 F; then 0.1 moles required 0.2 F.