Answer
$J = 1.1\times 10^6~A/m^2$
Work Step by Step
We can find the cross-sectional area where the current flows:
$A = (\pi)~(2.0\times 10^{-3}~m)^2 - (\pi)~(1.0\times 10^{-3}~m)^2$
$A = 9.42\times 10^{-6}~m^2$
We can find the current density:
$J = \frac{I}{A}$
$J = \frac{10~A}{9.42\times 10^{-6}~m^2}$
$J = 1.1\times 10^6~A/m^2$
