Answer
$8.79\times10^{-2}\;J$
Work Step by Step
The maximum kinetic energy associated with each wavelength of the wave is
$k_{max}=\frac{1}{4}\mu \omega^2 y_m^2\lambda$
or, $k_{max}=\frac{1}{4}\mu (2\pi f)^2 y_m^2\frac{v}{f}$
or, $k_{max}=\frac{1}{4}\mu (2\pi f)^2 y_m^2\frac{1}{f}\sqrt {\frac{T}{\mu}}$
or, $k_{max}=\pi^2 fy_m^2\sqrt {T\mu}$
As the standing wave is created by two waves traveling in opposite directions along the string, the maximum kinetic energy of the string in the loop during its oscillation is
$K_{max}=2k_{max}$
or, $K_{max}=2\pi^2 fy_m^2\sqrt {T\mu}$
Substituting the given values, we obtain
$K_{max}=2\times\pi^2 \times 120\times (5\times 10^{-3})^2\sqrt {40\times\frac{0.125}{2.25}}\; J$
or, $K_{max}=8.79\times10^{-2}\;J$
