Answer
$i = 18.1~\mu A$
Work Step by Step
We can assume that the inner part of the thin ring is $r = 1.195~mm$ from the center, and the outer part of the thin ring is $r = 1.205~mm$ from the center.
We can find the current in the thin ring:
$i = \int_{1.195~mm}^{1.205~mm}~(2\pi~r~dr)(Br)$
$i = \int_{1.195~mm}^{1.205~mm}~2\pi~B~r^2~dr$
$i = (\frac{2~\pi~B}{3})~r^3~\Big \vert_{1.195~mm}^{1.205~mm}$
$i = (\frac{2~\pi~B}{3})~[(1.205\times 10^{-3}~m)^3-(1.195\times 10^{-3}~m)^3]$
$i = \frac{(2~\pi)(2.00\times 10^5~A/m^3)}{3}~(4.32\times 10^{-11}~m^3)$
$i = 18.1\times 10^{-6}~A$
$i = 18.1~\mu A$
