Answer
The direction of the current is east.
Work Step by Step
Let east be the positive direction.
We can find the net current density:
$J = (ne)v_d$
$J = (2.80\times 10^{21}~m^{-3})(2)(1.6\times 10^{-19}~C)(25.0~m/s)+(2.80\times 10^{21}~m^{-3})(2)(-1.6\times 10^{-19}~C)(-88.0~m/s)$
$J = 1.01\times 10^5~A/m^2$
We let east be the positive direction, and the current density has a positive value.
Therefore, the direction of the current is east.
