Answer
See a detailed answer below.
Work Step by Step
a) We know that the net force exerted on the cylinder is zero since it is floating on rest.
$$\sum F_y=F_B-m_og=m_oa_y=m_o(0)=0$$
$$F_B=m_og$$
where the mass is given by density law $\rho=m/V$
Hence,
$$F_B=\rho_oV_og$$
And we know, from Archimedes’ principle, that the buoyant force is given by the weight of the displaced liquid.
$$m_{\rm f,displaced}\color{red}{\bf\not}g=\rho_oV_o\color{red}{\bf\not}g$$
$$\rho_{\rm f}V'=\rho_oV_o $$
where $V'$ here is the volume of the displaced liquid which is the volume of the submerged part of the cylinder under the liquid surface as well.
So,
$$\rho_{\rm f}V'_o =\rho_oV_o $$
Noting that $V_o=Al$, and $V'=Ah$
$$\rho_{\rm f}\color{red}{\bf\not}Ah=\rho_o\color{red}{\bf\not} Al$$
Hence,
$$\boxed{h= \dfrac{\rho_o l}{\rho_{\rm f}} }$$
_________________________________________________
b) Since the cylinder is at a distance $y$ above its equilibrium position, it must move down. This means that the net force exerted on it must be directed downward.
$$\sum F_y=F_B-m_og $$
By the same approach above;
$$\sum F_y=\rho_{\rm f}V'g-\rho_0V_og $$
Noting that $V_o=Al$, and $V'=A(h-y)$
$$\sum F_y=\rho_{\rm f}A(h-y)g-\rho_0Alg $$
$$\sum F_y=\rho_{\rm f}Ahg-\rho_{\rm f}A yg -\rho_0Alg $$
Noting that $\rho_{\rm f}Ahg=\rho_0Alg $ which is the weight of the displaced liquid is equal to the weight of the cylinder, as we found in part (a) above.
Thus,
$$\boxed{\sum F_y= -\rho_{\rm f}Ag y }$$
_________________________________________________
c) As the author told us we can recognize our result of part (b) as a version of
Hooke’s law, $F_{sp}=-kx$, and in our case, the distance is vertical as if we have a vertical spring.
So,
$$\sum F_y=F_{sp}$$
$$\color{red}{\bf\not}-\rho_{\rm f}Ag \color{red}{\bf\not} y=\color{red}{\bf\not}-k\color{red}{\bf\not}y$$
Hence,
$$\boxed{k=\rho_{\rm f}Ag}$$
_________________________________________________
d) The angular frequency for the simple harmonic motion of the cylinder when it is bobbing up and down is given by
$$\omega =2\pi f=\sqrt{\dfrac{k}{m_o}}$$
Recall that $f=1/T$, so
$$\dfrac{ 2\pi }{T}=\sqrt{\dfrac{k}{m_o}}$$
Hence,
$$T=2\pi \sqrt{\dfrac{m_o}{k}}$$
Plugging from above;
$$T=2\pi \sqrt{\dfrac{\rho_o \color{red}{\bf\not}Al}{\rho_{\rm f}\color{red}{\bf\not}Ag}}$$
$$T=2\pi \sqrt{\dfrac{\rho_o l}{\rho_{\rm f} g}}\tag 1$$
Solve the boxed formula in part (a) above for $l$ and plug it here.
$$T=2\pi \sqrt{\dfrac{\color{red}{\bf\not}\rho_o h\color{red}{\bf\not}\rho_{\rm f}}{\color{red}{\bf\not}\rho_{\rm f}\color{red}{\bf\not}\rho_o g}}$$
$$\boxed{T=2\pi \sqrt{\dfrac{h}{ g}}}$$
_________________________________________________
e) Plugging the known into (1);
$$T=2\pi \sqrt{\dfrac{ (917)(100)}{(1030) (9.8)}} =\color{red}{\bf 18.9} \;\rm s$$
