Answer
$300m/s$, in the negative y-direction.
Work Step by Step
We can find the magnitude and direction of the required velocity as follows:
$v=\frac{1}{B}(E-\frac{F_{net}}{q})$
We plug in the known values to obtain:
$v=\frac{1}{1.02T}(1250N/C-\frac{6.23\times 10^{-3}N}{6.6\times 10^{-9}C})$
$\implies v=300m/s$
According to the right-hand rule, we point the thumb in the direction of the net force, then point the fingers in the direction of the magnitude of the field, and curl the middle finger up to $90^{\circ}$. Thus, the direction of velocity is in the negative y direction.
