Answer:
P=3.42×10^-6 J/s
Explanation:
From the kinematics of motion with constant acceleration we know that :  
vf^2=vi^2+2*a(xf-xi)
Where : 
• vf , vi, are the the final and the initial velocity of the electron  
• a is the acceleration of the electron  
• xf , xi are the final and the initial position of the electron . 
Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  
Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  
 vf^2 =vi^2+2*a(xf-xi)
 vf^2-vi^2=2*a(xf-xi)
2*a(xf-xi)= vf^2-vi^2
           a = (vf^2-vi^2)/2(xf-xi)
Pluging known information to get : 
a = (vf^2-vi^2)/2(xf-xi)
   = 1.411 × 10^17
From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  
so,
 vf^2 =vi^2+2*a(xf-xi)
 vf^2 =5.312× 10^7
From the following Eq. we can calculate the time elapsed in this motion .  
 xf =xi+vi*t+1/2*a*t
 xf =xi+vi*t+1/2*a*t
   t=√2(xf-xi)/a
   t=3.765×10^-10 s
now we can use the power P Eq.  
  P=W/Δt => ΔK/Δt  
Where: the work done W change the kinetic energy K of the electron ,
ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2
P=1/2*m*vf^2-1/2*m*vi^2/Δt 
P=3.42×10^-6 J/s