Answer
$2.56\;\rm mA$
Work Step by Step
Since the positive ions move to the right while the negative ions move to the left, the net current is the sum of the two currents of both processes.
The direction of the current is in the positive ions direction.
Hence,
$$I_+=\dfrac{N_+q_+}{t}$$
and
$$I_-=-\dfrac{N_-q_-}{t}$$
$$I_{net}=I_++I_-$$
$$I_{net}=\dfrac{N_+q_+}{t}-\dfrac{N_-q_-}{t}$$
$$I_{net}= \dfrac{N_+q_+-N_-q_-}{t}$$
where $q_+=2e$ and $q_-=-e$
$$I_{net}= \dfrac{2N_+ e+N_-e}{t}$$
Plug the known;
$$I_{net}= \dfrac{2(5\times 10^{15})(1.6\times 10^{-19})+(6\times 10^{15})(1.6\times 10^{-19})}{1}$$
$$I=\bf 0.00256\;\rm A=\color{red}{\bf 2.56}\;\rm mA$$
