How many Au atoms can be extracted from 0.250 L of seawater?
You also need to convert the grams of gold into moles in order to use Avogadro's number.

Atomic weight of gold = $\pu{196.967 g/mol}$
$$
\begin{align}
\frac{\pu{1.5E-04 g/ton}}{\pu{196.967 g/mol}}&amp;= \pu{7.615E-07 mol/ton}\\
\pu{7.615E-07 mol/ton}\times \pu{6.022E+23 atoms/mol} &amp;= \pu{4.586E+17 atoms/ton}\\
\frac{\pu{4.586E{+17} atoms/ton}}{\pu{2000 lb/ton}}&amp;=\pu{2.293E{+14} atoms/lb}\\
\frac{\pu{2.293E{+14} atoms/lb}}{\pu{0.454 kg/lb}}=\pu{5.056E{+14} atoms/kg}&amp;=\pu{5.056E{+11} atoms/g}\\
\pu{5.056E{+11} atoms/g} \times \pu{1.03 g/ml}&amp;=\pu{5.208E{+11} atoms/ml}\\
\pu{5.208E{+11} atoms/ml}\times \pu{250 ml}&amp;= \pu{1.302 E{+14} atoms}/\pu{250 ml}\\
\end{align}
$$
There would be $1.302 \times10^{14}$ atoms of gold in $\pu{0.250 L}$ of seawater.