Answer
$2.3\;\rm mA$
Work Step by Step
We know that the current in the wire is given by
$$I=\dfrac{\Delta V_{\rm wire}}{R}$$
and since the wire is connected directly to the two ends of the battery, they have the same potential difference, $V_B=\Delta V_{\rm wire}$
$$I=\dfrac{ V_{\rm B}}{R}\tag 1$$
Now we need to find the resistance of the wire which is given by
$$R=\dfrac{\rho L}{A}=\dfrac{\rho L}{\pi r^2}$$
Plug into (1),
$$I=\dfrac{ \pi r^2V_{\rm B}}{\rho_{\rm gold} L} $$
$$I=\dfrac{ \pi (0.05\times 10^{-3})^2(0.7)}{(2.4\times 10^{-8})(100)} $$
$$I=\color{red}{\bf 2.3}\;\rm mA$$
