Answer
$ 22.6\;\rm mA$
Work Step by Step
We know that the current is given by
$$I=\dfrac{Q}{t}=\dfrac{Nq}{t}$$
where $N$ is the number of ions and $q=e$ is the charge of one gold ion.
$$I =\dfrac{Ne}{t}\tag 1$$
Now we need to find the amount of atoms (ions) in 0.5 g gold.
$$N=\dfrac{m}{M}N_{\rm A}$$
where $m$ is the mass of the gold piece while $M$ is the atomic number, and $N_{\rm A}$ is Avogadro's number.
Plug into (1),
$$I =\dfrac{ e}{t}\dfrac{m}{M}N_{\rm A}$$
Plug the known;
$$I =\dfrac{ (1.6\times 10^{-19})}{(3\times 60\times 60)}\dfrac{(0.5)}{197}(6.022\times 10^{23})$$
The author asked for the current in mA.
$$I=\color{red}{\bf 22.6}\;\rm mA$$
