Answer
$B = 0.495~T$
Work Step by Step
We can use the kinetic energy after accelerating through the potential difference to find the ion's speed:
$K = \vert q \vert~V_1$
$\frac{1}{2}mv^2 = \vert q \vert~V_1$
$v^2 = \frac{2~\vert q \vert~V_1}{m}$
$v = \sqrt{\frac{2~\vert q \vert~V_1}{m}}$
$v = \sqrt{\frac{(2)(3.20\times 10^{-19}~C)(1.00\times 10^5~V)}{3.92\times 10^{-25}~kg}}$
$v = 4.04\times 10^5~m/s$
We can use the equation for the radius of curvature to find the magnitude of the magnetic field:
$r = \frac{mv}{q~B}$
$B = \frac{mv}{q~r}$
$B = \frac{(3.92\times 10^{-25}~kg)(4.04\times 10^5~m/s)}{(3.20\times 10^{-19}~C)(1.00~m)}$
$B = 0.495~T$
