To calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2, we need to understand the concepts of conduction current and displacement current.

1. Conduction Current:
Conduction current is the flow of electric charge through a conductor due to the movement of free electrons. It is directly proportional to the electric field applied and inversely proportional to the conductivity of the material. The formula for conduction current is given by:
I_conduction = σ * A * E
where I_conduction is the conduction current, σ is the conductivity of the material, A is the cross-sectional area of the conductor, and E is the electric field.

2. Displacement Current:
Displacement current is a term introduced by Maxwell's equations to account for the changing electric field in a region. It is given by the formula:
I_displacement = ε * A * dE/dt
where I_displacement is the displacement current, ε is the permittivity of the material, A is the cross-sectional area of the region, and dE/dt is the rate of change of electric field with respect to time.

Now, we need to find the frequency at which the conduction and displacement currents become equal. Since the conductivity is unity and the permittivity is 2, we can equate the two currents and solve for frequency.

σ * A * E = ε * A * dE/dt
σ * E = ε * dE/dt
E/ε = dE/dt
Integrating both sides, we get:
∫ E/ε dE = ∫ dt
(E^2)/(2ε) = t + C
E^2 = 2ε(t + C)
E = √(2ε(t + C))

Since we are considering unity conductivity, σ = 1, and the electric field is directly proportional to the electric current. Therefore, we can write:
I = E

Substituting this in the equation, we get:
I = √(2ε(t + C))

Now, we can calculate the frequency at which the conduction and displacement currents become equal by finding the value of t when I = √(2ε(t + C)).

The correct answer is option 'B' - 9 GHz.