A wave with a frequency of 200 Hz and an amplitude of 1 mm travels...

Given:
Frequency, $f = 200 \text{ Hz}$
Amplitude, $A = 1 \text{ mm} = 10^{-3} \text{ m}$
Linear mass density, $\mu = 6 \text{ g m}^{-1} = 6 \times 10^{-3} \text{ kg m}^{-1}$
Length of the string, $L = 2.0 \text{ m}$
Tension, $T = 60 \text{ N}$

The wave speed is $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{60}{6 \times 10^{-3}}} = 100 \text{ m s}^{-1}$.
The wavelength is $\lambda = \dfrac{v}{f} = \dfrac{100}{200} = 0.5 \text{ m}$.
Since $L = 2.0 \text{ m} = 4\lambda$, the length is an integer multiple of the wavelength, so the average energy density formula can be used exactly.

The total energy $E$ of the wave in a length $L$ is given by:
$E = \dfrac{1}{2} \mu \omega^2 A^2 L$

Substituting $\omega = 2 \pi f$:
$E = \dfrac{1}{2} \mu (2 \pi f)^2 A^2 L$
$E = 2 \pi^2 \mu f^2 A^2 L$

Substituting the given values:
$E = 2 \pi^2 (6 \times 10^{-3}) (200)^2 (10^{-3})^2 (2.0)$
$E = 2 \pi^2 \times 6 \times 10^{-3} \times 40000 \times 10^{-6} \times 2.0$
$E = 9.6 \pi^2 \times 10^{-4} \text{ J}$

Using $\pi^2 \approx 9.87$:
$E \approx 9.6 \times 9.87 \times 10^{-4} \text{ J}$
$E \approx 94.75 \times 10^{-4} \text{ J}$
$E \approx 9.475 \times 10^{-3} \text{ J} \approx 9.5 \text{ mJ}$