Answer
$7.1*10^{-29} $ $ m$
Work Step by Step
Since $v = 5*10^6 m/s$ and $mv^2/r = 3.2 * 10^{-15}$, all we have to do is substitute values in to find r. $r$ is given by $\frac{m(5*10^6)^2}{3.2*10^{-15}}$. The particle has to be an electron, as the field is out of the page, and doing RH, we find that a proton would be traveling clockwise in a circle. Therefore, $m = 9.11 * 10^{-31}$ kg, so plugging in, we end up with $7.1*10^-29 m$
