1. For a wire, resistance is $R=\rho\frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

2. Given:

• Material $B$ has twice the resistivity of $A$: $\rho_B=2\rho_A$
• Diameter of wire $B$ is twice that of wire $A$: $d_B=2d_A$

3. Cross-sectional area of a circular wire is proportional to the square of diameter: $A\propto d^2$ So, $A_B=(2)^2A_A=4A_A$

4. For same resistance, $R_A=R_B$ Thus, $\rho_A\frac{L_A}{A_A}=\rho_B\frac{L_B}{A_B}$ Substitute $\rho_B=2\rho_A$ and $A_B=4A_A$: $\rho_A\frac{L_A}{A_A}=2\rho_A\frac{L_B}{4A_A}$

5. Cancel common terms $\rho_A$ and $A_A$: $L_A=\frac{2L_B}{4}=\frac{L_B}{2}$ Hence, $L_B=2L_A$ So, $\frac{L_B}{L_A}=2$

6. Therefore, the correct option is: $\boxed{2}$ which is option $\boxed{D}$.

Comparison with stored answer: Stored correct answer is $D$, which matches the derived answer.