The correct answer is $\boxed{\text{A}}$.
The specific resistance of a material is the resistance of a wire of that material with a unit length and unit cross-sectional area. It is denoted by $\rho$.
The resistance of a wire is given by the equation $R = \rho \frac{l}{A}$, where $l$ is the length of the wire, $A$ is the cross-sectional area of the wire, and $\rho$ is the specific resistance of the material.
In this case, we are given that $l = 200 \text{ m}$, $R = 21 \Omega$, and $d = 0.44 \text{ mm}$. The diameter of the wire is $d = 2r = 0.44 \times 2 = 0.88 \text{ mm}$. The cross-sectional area of the wire is $A = \pi r^2 = \pi \left(0.44 \times 10^{-3} \text{ m}\right)^2 = 1.3 \times 10^{-6} \text{ m}^2$.
Substituting these values into the equation for resistance, we get $R = \rho \frac{l}{A} = 21 \Omega = \rho \frac{200 \text{ m}}{1.3 \times 10^{-6} \text{ m}^2}$. Solving for $\rho$, we get $\rho = \frac{21 \Omega \times 1.3 \times 10^{-6} \text{ m}^2}{200 \text{ m}} = 1.2 \times 10^{-8} \Omega \text{ m}$.
Therefore, the specific resistance of the copper wire is $\boxed{1.2 \times 10^{-8} \Omega \text{ m}}$.
The other options are incorrect because they are not equal to the specific resistance of the copper wire.