Answer
$v = 300~m/s$
Work Step by Step
Let $A$ be the cross-sectional area of the wire.
We can find an expression for the maximum tension:
$\tau = (7.00\times 10^8~N/m^2)(A)$
We can find an expression for the linear density:
$\mu = \frac{m}{L} = \frac{V~\rho}{L} = \frac{A~L~\rho}{L} = A~\rho$
We can find the maximum wave speed:
$v = \sqrt{\frac{\tau}{\mu}}$
$v = \sqrt{\frac{(7.00\times 10^8~N/m^2)(A)}{A~\rho}}$
$v = \sqrt{\frac{7.00\times 10^8~N/m^2}{\rho}}$
$v = \sqrt{\frac{7.00\times 10^8~N/m^2}{7800~kg/m^3}}$
$v = 300~m/s$
