Emulate impedance response using RLC components
The expression for impedance with one inductor and two resistors is
\$
R1 + j\omega\ R2L1 / (R2 + j\omega\ L1)
\$
At low frequency the impedance is \$\approx R1\$ and at very high frequency the impedance is \$\approx\ R1 + R2\$. With the three tunable parameters (R1,R2,L1) you can match only the low frequency impedance, high frequency impedance and the corner frequency.
As @a concerned citizen mentioned in the comment, you may need more components for closer match.
The mismatch at 2kHz can be bridged by adding another impedance in series which has low values at frequencies below and above 2kHz. A parallel LC circuit can be used for this purpose. A sample circuit with sample values (not tuned for best match) is shown below. LC was chosen so that
\$2*\pi* 2kHz \approx 1/\sqrt{LC}\$.
The parallel LC circuit will have infinite impedance at 2kHz. However, we want to bridge only the 5ohms difference seen in your graph. Hence, a 5ohm resistor is also kept in parallel. The above calculations are all thumb rules / approximate formulae.



simulate this circuit &ndash; Schematic created using CircuitLab
The plot of the R|L|C section's impedance is shown below.

The net impedance is shown below.

You can independently verify if the given scheme is suitable. For a better match, use the tuning method you have used for the calculation of (R1, R2, L1).