Given Information:
- Conductivity of the material: 10-2 mho/m
- Relative permittivity of the material: 10

Concepts:
- Conduction current: The current flowing through a material due to the movement of charge carriers.
- Displacement current: The current that arises in a region where the electric field is changing with time.
- The conduction current is given by the equation: Ic = σE, where Ic is the conduction current, σ is the conductivity, and E is the electric field.
- The displacement current is given by the equation: Id = ε0εr(dE/dt), where Id is the displacement current, ε0 is the permittivity of free space, εr is the relative permittivity, and dE/dt is the rate of change of electric field.

Approach:
We need to find the frequency at which the conduction current is equal to the displacement current. To do this, we'll equate the two currents and solve for the frequency.

Solution:
1. Equate the conduction current and the displacement current equations:
σE = ε0εr(dE/dt)

2. Rearrange the equation to solve for dE/dt:
dE/dt = (σ/ε0εr)E

3. The rate of change of electric field can be expressed as:
dE/dt = jωE, where j is the imaginary unit and ω is the angular frequency.

4. Substitute this expression into the equation obtained in step 2:
jωE = (σ/ε0εr)E

5. Cancel out E on both sides of the equation:
jω = (σ/ε0εr)

6. Solve for ω:
ω = (σ/ε0εr)/j

7. Substitute the given values of σ and εr:
ω = (10-2 mho/m)/(8.85 x 10-12 F/m x 10) / j

8. Simplify the expression:
ω = (10-2 mho/m)/(8.85 x 10-11 F/m) / j

9. Convert mho to siemens (S):
ω = (10-2 S/m)/(8.85 x 10-11 F/m) / j

10. Simplify further:
ω = (10-2 S/m x 1/8.85 x 10-11 F/m) / j
= 1.1299 x 109 / j

11. Calculate the magnitude of the angular frequency:
|ω| = 1.1299 x 109 rad/s

12. Convert the angular frequency to frequency in MHz:
f = |ω| / (2π x 106)
≈ 18 MHz

Answer:
The frequency at which the conduction current in the medium is equal to the displacement current is approximately 18 MHz.

Given :σ = 10² mho/m