Step 1: Write the relation between resistance and temperature.
The resistance of a conductor at temperature $T$ is given by the relation:
\[
R = R_0 \left(1 + \alpha (T - T_0)\right)
\]
where $R_0$ is the resistance at reference temperature $T_0$, and $\alpha$ is the temperature coefficient of resistance.
Step 2: Substitute the given condition.
It is given that the resistance becomes twice the resistance at $20^\circ$C.
Therefore,
\[
R = 2R_0
\]
Substituting in the formula:
\[
2R_0 = R_0 \left(1 + \alpha (T - 20)\right)
\]
Dividing both sides by $R_0$:
\[
2 = 1 + \alpha (T - 20)
\]
Step 3: Substitute the value of $\alpha$.
\[
2 = 1 + (4.0 \times 10^{-3})(T - 20)
\]
\[
1 = (4.0 \times 10^{-3})(T - 20)
\]
Step 4: Solve for temperature.
\[
T - 20 = \frac{1}{4.0 \times 10^{-3}}
\]
\[
T - 20 = 250
\]
\[
T = 270^\circ C
\]
Step 5: Final answer.
The resistance of the silver wire becomes twice its value at
\[
T = 270^\circ C
\]