Let's take the x,y,z axis such that the electric field E and the magnetic field B are on the x-axis and the velocity of the particle is on the y-axis.
The charge is [tex]q=1.8 \mu C=1.8 \cdot 10^{-6} C[/tex].
The electric force acting on the charge is
[tex]F_E = qE=(1.6 \cdot 10^{-6} C)(5.7 \cdot 10^3 N/C)=9.1 \cdot 10^{-3} N[/tex]
and the direction is the same as the electric field, so on the x-axis.
The magnetic force acting on the charge is
[tex]F_B = qvB = (1.6 \cdot 10^{-6} C)(3 \cdot 10^6 m/s)(1.2 \cdot 10^{-3}T)=5.8 \cdot 10^{-3}N[/tex]
and by using the right hand rule, we find that the direction of the force is on the z-axis.
So, the two forces Fe and Fb are perpendicular to each other. Therefore, the net force acting on the particle is the resultant of the two forces:
[tex]F= \sqrt{F_E^2+F_B^2}= \sqrt{(9.1 \cdot 10^{-3} N)^2+(5.8 \cdot 10^{-3}N)^2} =0.011 N[/tex]
