Answer
The answer does not depend on the diameter of the wire.
Work Step by Step
Let $A$ be the cross-sectional area of the wire.
We can find an expression for the maximum tension:
$\tau = (7.00\times 10^8~N/m^2)(A) = (7.00\times 10^8~N/m^2)(\pi~r^2) = (7.00\times 10^8~N/m^2)(\pi)~(\frac{d}{2})^2$
We can find an expression for the linear density:
$\mu = \frac{m}{L} = \frac{V~\rho}{L} = \frac{A~L~\rho}{L} = A~\rho = \pi~r^2~\rho = \pi~(\frac{d}{2})^2~\rho$
We can see that $\tau \propto d^2$ and $\mu \propto d^2$
We can write an expression for the maximum wave speed:
$v = \sqrt{\frac{\tau}{\mu}}$
$v = \sqrt{\frac{(7.00\times 10^8~N/m^2)(\pi)~(\frac{d}{2})^2}{\pi~(\frac{d}{2})^2~\rho}}$
$v = \sqrt{\frac{(7.00\times 10^8~N/m^2)}{\rho}}$
In this expression, the $d^2$ cancels out of the numerator and the denominator.
Therefore, the answer does not depend on the diameter of the wire.
