Answer
(a) $1.1\frac{m}{s}$
(b) bottom electrode; no
Work Step by Step
(a) We know that
$v=\frac{E}{B}$
We plug in the known values to obtain:
$v=\frac{195\times 10^{-6}}{(2.75\times 10^{-3})(0.065)}$
$v=1.1\frac{m}{s}$
(b) As we know that the electric field is directed towards the decreasing potential, thus the bottom electrode is at the higher potential and our answer does not depend on the sign of the ions.
