Answer
$J=6.4\frac{A}{m^2}$
Work Step by Step
The current density is given as
$J=nqv_d$ ...........eq(1)
But given that ions are doubly charged, we find:
$q=2e=2\times1.6\times10^{-19}$
$n=2.0\times\frac{10^8}{cm^3}=2\times\frac{10^8}{(10^{-2}m)^3}=2\times10^8\times10^6m^3=2\times10^{14}m^3$
We plug in the known values to obtain:
$J=2\times10^{14}\times2\times1.6\times10^{-19}\times1\times10^5$
$J=6.4\frac{A}{m^2}$
