Step 1: General result:
For a fixed wire length $\ell$, $N=\dfrac{\ell}{\pi D}$ and $A=\dfrac{\pi D^2}4$, so $m=I\dfrac{\ell}{\pi D}\cdot\dfrac{\pi D^2}{4}=\dfrac{I\ell D}4$.
Step 2: Dependence:
So $m\propto D$ for the same wire and current.
Step 3: Ratio:
Doubling the diameter doubles the moment. Ratio $2:1$. Option (A).
Final Answer:
For a fixed wire length, m is proportional to the diameter.
\[ \boxed{A} \]