The reduction reaction for calcium from molten $\text{CaCl}_2$ is: $\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}$ This indicates that $1 \text{ mole}$ of calcium ($40 \text{ g}$) requires $2 \text{ moles}$ of electrons, which corresponds to $2 \text{ Faradays (F)}$ of charge . Given mass of calcium to be produced = $20 \text{ g}$. Number of moles of calcium = $\frac{\text{Given mass}}{\text{Atomic mass}} = \frac{20 \text{ g}}{40 \text{ g mol}^{-1}} = 0.5 \text{ mol}$. Since $1 \text{ mol}$ of Ca requires $2 \text{ F}$, Charge required for $0.5 \text{ mol}$ of Ca = $0.5 \times 2 \text{ F} = 1 \text{ F}$.