= Solution
Using $M_{\rm Au}\simeq196.97\ {\rm g\,mol^{-1}}$ and the <Avogadro constant>, the areal atom count is
$$
n_{\rm Au}=\frac{1.56\times10^{-6}\ {\rm g}}{196.97\ {\rm g\,mol^{-1}}}
\frac{6.02214\times10^{23}\ {\rm mol^{-1}}}{10^{-4}\ {\rm m^2}}
\simeq4.77\times10^{19}\ {\rm m^{-2}}.
$$
One atom occupies each square cell of area $a^2$, so the inferred <square lattice> spacing is
$$
\boxed{a=n_{\rm Au}^{-1/2}\simeq1.45\times10^{-10}\ {\rm m}=0.145\ {\rm nm}.}
$$
This uses the given deposited <mass> and the assumed single square layer; it does not substitute a bulk <gold> lattice parameter for the measured areal count.
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