Step 1: Replace the hollow tube with an equivalent solid wire of the same cross-section area.
Because resistance depends on area only through $R = \rho L/A$, a hollow tube behaves exactly like a solid wire whose area matches the tube's ring-shaped area. If $d_{eq}$ is the diameter of that equivalent solid wire, then $d_{eq}^2 = d_{outer}^2 - d_{inner}^2$, since area scales as diameter squared and the ring's area is the outer disc's area minus the inner disc's area.
Step 2: Compute this equivalent diameter for tube B.
\[ d_{eq}^2 = (2.0)^2 - (1.0)^2 = 4 - 1 = 3\ \text{mm}^2 \] So $d_{eq} = \sqrt{3}$ mm.
Step 3: Compare two solid wires (A and B's equivalent) of the same material and length.
\[ \frac{R_1}{R_2} = \left(\frac{d_{eq}}{d_1}\right)^2 \]
Step 4: Substitute the given ratio.
\[ 3 = \left(\frac{\sqrt{3}}{d_1}\right)^2 = \frac{3}{d_1^2} \]
Step 5: Solve for $d_1$.
\[ d_1^2 = 1 \quad\Rightarrow\quad d_1 = 1.0\ \text{mm} \]
Final Answer:
\[ \boxed{1.0\ \text{mm}} \]