Answer
$v = 2.8\times 10^7~m/s$
Work Step by Step
$F = \frac{mv^2}{r}$
$F = qvB$
We can equate the two expressions for $F$ to find the maximum possible value for $v$:
$\frac{mv^2}{r} = qvB$
$v = \frac{q~B~r}{m}$
$v = \frac{(1.6\times 10^{-19}~C)(0.360~T)(0.820~m)}{1.67\times 10^{-27}~kg}$
$v = 2.8\times 10^7~m/s$
