= Solution
Work in the small-slope model of a <transverse wave> on a taut string. Let $T$ be the constant <tension>, $\rho$ the <linear mass density>, and take a right-moving sinusoidal <displacement> $y=A\cos(\kappa x-\omega t)$, with <wave number> $\kappa=\omega/v$. The <energy flux of a transverse string wave> through a cut is the transverse <force> times transverse <velocity>:
$$
P(x,t)=-T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}
=T\kappa\omega A^2\sin^2(\kappa x-\omega t).
$$
A cycle average gives $\langle\sin^2\rangle=1/2$. Using $T=\rho v^2$, obtained from the string <wave equation>, yields
$$
\boxed{\overline P=\frac12T\kappa\omega A^2=\frac12\rho v\omega^2A^2.}
$$
The sign in the flux formula makes the <power> positive for propagation in the positive $x$ direction.
The <energy> per unit equilibrium length is the sum of the kinetic and elastic terms:
$$
\mathcal E=\frac12\rho y_t^2+\frac12T y_x^2,
\qquad \langle\mathcal E\rangle=\frac12\rho\omega^2A^2.
$$
For a segment of <mass> $M=\rho\ell$ containing many <wavelengths>, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq\tfrac12M\omega^2A^2$. Convert the given quantities to SI units: $M=0.080\ \mathrm{kg}$, $A=1.6\times10^{-4}\ \mathrm m$, and $\omega=2\pi(120)\ \mathrm{s}^{-1}$. Then
$$
\boxed{E\simeq\frac12(0.080)(240\pi)^2(1.6\times10^{-4})^2
=5.82\times10^{-4}\ \mathrm J.}
$$
The many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of <wavelengths>.
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