Answer
The amplitude must be 4.10 mm
Work Step by Step
We can find the angular frequency.
$\omega = 2\pi~f$
$\omega = (2\pi)~(69.0~Hz)$
$\omega = 433.5~rad/s$
We can find the linear mass density of the wire.
$v = \sqrt{\frac{F_T}{\mu}}$
$\mu = \frac{F_T}{v^2}$
$\mu = \frac{94.0~N}{(406~m/s)^2}$
$\mu = 5.70\times 10^{-4}~kg/m$
We can find the required amplitude.
$P_{av} = \frac{1}{2}\sqrt{\mu~F_T}~\omega^2~A^2$
$A^2 = \frac{2~P_{av}}{\sqrt{\mu~F_T}~\omega^2}$
$A = \sqrt{\frac{2~P_{av}}{\sqrt{\mu~F_T}~\omega^2}}$
$A = \sqrt{\frac{(2)(0.365~W)}{\sqrt{(5.70\times 10^{-4}~kg/m)(94.0~N)}~(433.5~rad/s)^2}}$
$A = 0.00410~m$
$A = 4.10~mm$
The amplitude must be 4.10 mm
