Answer
The singly charged ion has a path of larger radius by a factor of 2.
Work Step by Step
We can find an expression for the radius of the path:
$F = \frac{mv^2}{r}$
$qvB = \frac{mv^2}{r}$
$r = \frac{mv^2}{qvB}$
$r = \frac{mv}{qB}$
We can see that the doubling the charge $q$ will result in a radius that is half the original radius. Therefore, the singly charged ion has a path of larger radius by a factor of 2.
