Since charge is moving inside both type of field
there will exist both forces on it
now
[tex]F = q(v x B) + qE[/tex]
If net force on moving charge is zero
[tex]0 = qvBsin90 + qE[/tex]
[tex]qvB = qE[/tex]
hence we will have speed
[tex]v = \frac{E}{B}[/tex]
[tex]v = \frac{9 * 10^5}{1.3}[/tex]
[tex]v = 6.92 * 10^5 m/s[/tex]