Answer
$B = 0.17~T$
Work Step by Step
$F = \frac{mv^2}{r}$
$F = qvB$
We can equate the two expressions for $F$ to find an expression for the speed $v$:
$\frac{mv^2}{r} = qvB$
$v = \frac{q~B~r}{m}$
The ion has a kinetic energy of $q~\Delta V$. We can find the magnitude of the field:
$K = q~\Delta V$
$\frac{1}{2}mv^2 = q~\Delta V$
$\frac{1}{2}(m)(\frac{q~B~r}{m})^2 = q~\Delta V$
$B^2 = \frac{2~m~\Delta V}{q~r^2}$
$B = \sqrt{\frac{2~m~\Delta V}{q~r^2}}$
$B = \sqrt{\frac{(2)~(12)(1.66\times 10^{-27}~kg)~(5000~V)}{(1.6\times 10^{-19}~C)~(0.21~m)^2}}$
$B = 0.17~T$
