The Wurtz reaction between methyl chloride and sodium metal in presence of dry ether can be written as:
$2C{H}_3+Cl+2Na\stackrel{\mathrm{D}\mathrm{r}\mathrm{y}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}}{\to }C{H}_3\underset{\mathrm{E}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{e}}{-}C{H}_3+2NaCl$
$\Rightarrow$ moles of sodium metal reacts to give 1 mole of ethane.
Also,
1 mole of Na metal $=23g$
$\therefore$ 2 moles of Na metal $=23\times 2=46g$
Thus, 46 g of sodium is required to prepare one mole of ethane from methyl chloride by Wurtz reaction.