Answer
$i = 0.666~A$
Work Step by Step
We can find the current:
$i = \int_{0}^{R}~J_0(1-\frac{r}{R})(2\pi~r~dr)$
$i = \int_{0}^{R}~2\pi~J_0~(r-\frac{r^2}{R})~dr$
$i = 2\pi~J_0~(\frac{r^2}{2}-\frac{r^3}{3R})\Big \vert_{0}^{R}$
$i = 2\pi~J_0~(\frac{R^2}{2}-\frac{R^3}{3R})-0$
$i = 2\pi~J_0~(\frac{R^2}{2}-\frac{R^2}{3})$
$i = 2\pi~J_0~(\frac{R^2}{6})$
$i = (2\pi)~(5.50\times 10^4~A/m^2)~\frac{(3.40\times 10^{-3}~m)^2}{6}$
$i = 0.666~A$
