When a `40 Vm^(-1)` electric field produced inside a conductor then `2 xx 10^(4) Am^(-2)` curren density is established in it. Resistivity of the conducot is
(1) `2 xx 10^(-3) Omega m` (2) `4 xx 10^(-3) Omega m ` (3) ` 2 xx 10^(3)Omega m ` (4) ` 4 xx 10^(3) Omega m`

Explanation

To find the resistivity of the conductor given the electric field and current density, we can use the relationship between electric field (E), current density (J), and resistivity (ρ) as derived from Ohm's law.

### Step-by-Step Solution:

1. **Understand the relationship**: According to Ohm's law in the context of a conductor, the current density (J) is related to the electric field (E) and resistivity (ρ) by the formula:
\[ J = \frac{E}{\rho} \]
Rearranging this gives us:
\[ \rho = \frac{E}{J} \]

2. **Identify the given values**: From the problem, we have:
- Electric field, \( E = 40 \, \text{Vm}^{-1} \)
- Current density, \( J = 2 \times 10^{4} \, \text{Am}^{-2} \)

3. **Substitute the values into the formula**: Now, we can substitute the values of E and J into the rearranged formula for resistivity:
\[ \rho = \frac{40 \, \text{Vm}^{-1}}{2 \times 10^{4} \, \text{Am}^{-2}} \]

4. **Calculate the resistivity**: Performing the division:
\[ \rho = \frac{40}{2 \times 10^{4}} = \frac{40}{20000} = 0.002 \, \Omega \cdot m \]
This can also be expressed as:
\[ \rho = 2 \times 10^{-3} \, \Omega \cdot m \]

5. **Identify the correct option**: The calculated resistivity is \( 2 \times 10^{-3} \, \Omega \cdot m \), which corresponds to option (1).

### Final Answer:
The resistivity of the conductor is \( 2 \times 10^{-3} \, \Omega \cdot m \).