1. Identify the cathode reaction

At the cathode, gold is deposited from $\mathrm{AuCl_4^-}$:

$\mathrm{AuCl_4^- + 3e^- \rightarrow Au + 4Cl^-}$

So, 1 mole of Au requires 3 moles of electrons.

2. Calculate moles of gold deposited

Given mass of Au deposited = $1.314\,\mathrm{g}$

Molar mass of Au = $197\,\mathrm{g\,mol^{-1}}$

$n(\mathrm{Au}) = \frac{1.314}{197}$

$n(\mathrm{Au}) = 0.00667\,\mathrm{mol}$

3. Calculate moles of electrons passed

Since 1 mol Au needs 3 mol $e^-$,

$n(e^-) = 3 \times 0.00667 = 0.0200\,\mathrm{mol}$

4. Convert to faradays

1 faraday = 1 mole of electrons.

Therefore, total charge passed = $0.0200\,\mathrm{F}$

$0.0200\,\mathrm{F} = 2.00 \times 10^{-2}\,\mathrm{F}$

5. Fill in the blank

The question asks for

$\underline{\hspace{2cm}} \times 10^{-2}\,\mathrm{F}$

So the required integer is:

$\boxed{2}$