Answer
$\sqrt {2}$
Work Step by Step
Recall that the wave speed on a stretched wire is given by the relation,
$v=\sqrt {\frac{\tau}{\mu}}$ where $\tau$ is the tension in the wire and $\mu$ is the linear density of the wire.
Let the new wave speed be $v_{2}$ and the old wave speed be $v_{1}$.
Two conditions are given: $\tau_{2}=2\tau_{1}$, Linear density is the same in both cases.
Therefore the ratio of the new to the old wave speed is,
$\frac{v_{2}}{v_{1}}=\frac{\sqrt \frac{\tau_{2}}{\mu}}{\sqrt \frac{\tau_{1}}{\mu}}=\sqrt {\frac{\tau_{2}}{\tau_{1}}}=\sqrt {\frac{2\tau_{1}}{\tau_{1}}}=\sqrt {2}$
