The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
[tex]F=qE[/tex]
where in our problem [tex]q=3 \mu C= 3 \cdot 10^{-6} C[/tex] and [tex]E=2 kN/C=2000 N/C[/tex], so the force is
[tex]F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N[/tex]
The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
[tex]K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J[/tex]
