Answer:
The magnetic field required required for the beam not to be deflected  is [tex]B = 0.0036T[/tex]
Explanation:
From the question we are told that
  The charge on the particle is [tex]q = +2e[/tex]
  The mass of the particle is  [tex]m = 6.64 *10^{-27} kg[/tex]
  The potential difference is [tex]V_a = 1.8 kV = 1.8 *10^{3} V[/tex]
  The potential difference between the two parallel plate is  [tex]V_b = 120 V[/tex]
  The separation between the plate is  [tex]d = 8 mm = \frac{8}{1000} = 8*10^{-3}m[/tex]
 Â
The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam  after the region having a potential difference of 1.8kV
        [tex]KE_b = PE_b[/tex]
Generelly
       [tex]KE_b = \frac{1}{2} m v^2[/tex]
And    [tex]PE_b = q V_a[/tex]
 Equating this two formulas
       [tex]\frac{1}{2} mv^2 = q V_a[/tex]
making v the subject
      [tex]v = \sqrt{\frac{q V_a}{2 m} }[/tex]
Substituting value Â
      [tex]v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }[/tex]
      [tex]v = 41.65*10^4 m/s[/tex]
Generally the electric field between the plates is mathematically represented as
         [tex]E = \frac{V_b}{d}[/tex] Â
Substituting value Â
         [tex]E = \frac{120}{8*10^{-3}}[/tex]       Â
        [tex]E = 15 *10^3 NC^{-1}[/tex]
the magnetic field  is mathematically evaluate  Â
           [tex]B = \frac{E}{v}[/tex]
          [tex]B = \frac{15 *10^{3}}{41.65 *10^4}[/tex]
          [tex]B = 0.0036T[/tex]
