The half-cell reaction is,

${\mathrm{Au}}^{3+} + 3{\mathrm{e}}^{-} \to \mathrm{Au}$

Given W = $1.314 \mathrm{g}$

Atomic mass of $\mathrm{Au}=197$

When one Faraday of charge is passed through a circuit, then one equivalent of charge is deposited at the respective electrode.

$\frac{\mathrm{W}}{\mathrm{E}}=\frac{ \mathrm{charge}}{1 \mathrm{F}}$

$\frac{1.314}{\frac{197}{3}}=\frac{\mathrm{Q}}{1\mathrm{F}}$

$\mathrm{Q}=2\times {10}^{-2} \mathrm{F}$