1. Let the original wire be uniform

   Since the original wire is of the same material and same cross-sectional area throughout, resistance is proportional to length: $R = \rho \frac{L}{A}$

   Let the lengths of the two parts be: $X = x, \qquad Y = y$ with $x + y = 1\text{ m}$

2. Resistance of part $Y$

   If the original wire has resistivity $\rho$ and cross-sectional area $A$, then $R_Y = \rho \frac{y}{A}$

3. Part $X$ is stretched to form wire $W$

   Given:

   • Length of $W$ is twice the length of $X$: $L_W = 2x$

   When a wire is stretched, its volume remains constant: $A_X \cdot x = A_W \cdot 2x$ so $A_W = \frac{A}{2}$

4. Resistance of stretched wire $W$

   Resistance of $W$: $R_W = \rho \frac{L_W}{A_W} = \rho \frac{2x}{A/2} = \rho \frac{4x}{A}$

5. Use the given condition

   It is given that resistance of $W$ is twice that of $Y$: $R_W = 2R_Y$

   Substitute the expressions: $\rho \frac{4x}{A} = 2\left(\rho \frac{y}{A}\right)$

   Cancel $\rho/A$: $4x = 2y$ $2x = y$

6. Find the ratio $X:Y$

   $x:y = 1:2$

7. Check options

   • A: $1:4$ ❌
   • B: $1:2$ ✅
   • C: $4:1$ ❌
   • D: $2:1$ ❌

Therefore, the correct answer is: $\boxed{1:2}$