Answer
${\bf 1.155} \;\rm \mu A$
Work Step by Step
We know that the current is given by
$$I=\dfrac{Q}{t} $$
where $Q=eN$ where $e$ is the electron's charge and $N$ is the number of the electrons.
$$I=\dfrac{Ne}{t}\tag 1 $$
Since we assumed that there is no loss of energy by radiation or other means, the whole energy of the electrons will be transferred to the foil.
Thus, the change in the foil's thermal energy is given by
$$\Delta E_{\rm th}=mc\Delta T\tag 2$$
Now we can find the number of electrons since we know that each electron accelerates from $v$ to zero speed when it hits the foil.
Thus,
$$\Delta E_{\rm th}=N|\Delta K_{e}|=N|(0-eV)|$$
Hence,
$$N=\dfrac{\Delta E_{\rm th}}{e V}$$
Plug into (1);
$$I=\dfrac{\color{red}{\bf\not} e}{t}\dfrac{\Delta E_{\rm th}}{\color{red}{\bf\not} e V} $$
$$I= \dfrac{\Delta E_{\rm th}}{ t V} $$
Plug from (2);
$$I= \dfrac{mc\Delta T}{ t V} $$
Plug the known;
$$I= \dfrac{(10\times 10^{-6})(385)(6)}{ (10)(2000)} $$
$$I=\color{red}{\bf 1.155} \;\rm \mu A$$
