Answer
$i = 1.33~A$
Work Step by Step
We can find the current:
$i = \int_{0}^{R}~(\frac{J_0~r}{R})(2\pi~r~dr)$
$i = \int_{0}^{R}~\frac{2\pi~J_0~r^2}{R}~dr$
$i = \frac{2\pi~J_0~r^3}{3R}\Big \vert_{0}^{R}$
$i = \frac{2\pi~J_0~R^3}{3R}-0$
$i = \frac{2\pi~J_0~R^2}{3}$
$i = \frac{(2\pi)~(5.50\times 10^4~A/m^2)~(3.40\times 10^{-3}~m)^2}{3}$
$i = 1.33~A$
