A

To find the resistance of the wire, we first need to use the information given about the capacitor and the time constant in an RC circuit.

1. **Initial Data**:
- Capacitance, C = 2.0 µF = 2.0 \times 10^{-6} F
- Time, t = 4.4 µs = 4.4 \times 10^{-6} s
- The electric field drops to one third of its initial value.

2. **Relationship**: The voltage across the capacitor as it discharges can be modeled by the equation:
$$ V(t) = V_0 e^{-t/RC} $$
where V(t) is the voltage at time t, V_0 is the initial voltage, R is the resistance, and C is the capacitance.

3. **Setting Up the Ratio**: Since the electric field is proportional to voltage, when the voltage drops to \frac{1}{3} V_0, we have:
$$ V_0 e^{-t/RC} = \frac{1}{3} V_0 $$
This simplifies to:
$$ e^{-t/RC} = \frac{1}{3} $$

4. **Taking the Natural Logarithm**: Taking the natural logarithm on both sides gives:
$$ -\frac{t}{RC} = \ln(\frac{1}{3}) $$

5. **Rearranging**: Rearranging this gives us:
$$ R = -\frac{t}{C \ln(\frac{1}{3})} $$

6. **Substituting Values**: Now substituting in the known values:
$$ R = -\frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \ln(\frac{1}{3})} $$
The value of \ln(\frac{1}{3}) = -1.0986, therefore:
$$ R = -\frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6}\times(-1.0986)} $$

7. **Calculating R**:
$$ R = \frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \times 1.0986} $$
$$ = \frac{4.4}{2.0 \times 1.0986} \times 10^{0} \approx 2.00 \, \Omega $$

Therefore, the resistance of the wire is approximately 2.00 Ω.