Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the frequency noticed by an observer when the observer is moving toward the source while the source is stationary is given by
$$f'=\dfrac{v+v_o}{v}f_0\tag 1$$
where $v$ is the sound wave speed, $v_o$ is the speed of the observer, and $f$ is the original frequency at the source.
Now the observer is reflecting the sound wave and is now working as a new source that is moving toward a stationary observer which is the original source.
Hence, the frequency
$$f''=\dfrac{v}{v-v_o}f'$$
where $f''$ is the echo frequency, so plugging from (1)
$$f_{\rm echo}=\dfrac{v}{v-v_o}\dfrac{v+v_o}{v}f_0 $$
$$\boxed{f_{\rm echo}=\left[\dfrac{v+v_o}{v-v_o}\right] f_0 }$$
$$\color{blue}{\bf [b]}$$
The beat frequency is given by
$$f_{beat}=f_{echo}-f_0$$
$$f_{beat}=\left[\dfrac{v+v_o}{v-v_o}\right] f_0-f_0$$
$$f_{beat}=\left(\dfrac{v+v_o}{v-v_o} -1\right)f_0$$
$$f_{beat}=\left(\dfrac{1+\dfrac{v_o}{v}}{1-\dfrac{v_o}{v}} -1\right)f_0$$
Let's assume that $(x=v_o/v)$ where $(x\lt \lt 1)$ since $(v_o\lt\lt v)$
$$f_{beat}=\left(\dfrac{1+x}{1-x} -1\right)f_0$$
$$f_{beat}=\left(\dfrac{1+x-1+x}{1-x} \right)f_0=2xf_0(1-x)^{-1}$$
Using the binomial approximation for $(1-x)^{-1}\approx 1+x$
$$f_{beat} =2xf_0(1+x) $$
Thus,
$$f_{beat} =2f_0\dfrac{v_o}{v}\left(1+\dfrac{v_o}{v}\right) $$
$$f_{beat} =2f_0\left(\dfrac{v_o}{v} +\left[\dfrac{v_o}{v}\right]^2\right) $$
where $\left[\dfrac{v_o}{v}\right]^2\approx 0$ since $v_o\lt\lt v$
$$\boxed{f_{beat} =\left(\dfrac{2v_o}{v} \right)f_0 }$$
$$\color{blue}{\bf [c]}$$
We just need to plug the given into the previous boxed formula we derived in part [b] above. So we need to solve it for $v_0$;
$$ \dfrac{f_{beat}}{2f_0} = \dfrac{v_o}{v} $$
$$ v_o=\dfrac{f_{beat}}{2f_0}v = \dfrac{65}{2(2.4\times 10^6)}(1540)$$
$$v_o=\bf 0.0209\;\rm m/s=\color{red}{\bf2.09}\;\rm cm/s$$
$$\color{blue}{\bf [d]}$$
We know that $v_{o}=v_{\rm max}=A\omega$ where $\omega=2\pi f$
So,
$$v_o=2\pi f A$$
$$A=\dfrac{v_o}{2\pi f}$$
Plugging the known;
$$A=\dfrac{0.0209}{2\pi (\frac{90}{60})}=\bf 0.00221756\;\rm m$$
$$A=\color{red}{\bf2.22}\;\rm mm$$
