Answer
$ \dfrac{h }{l } =\sqrt[3]{1-\dfrac{ \rho_{\rm o}}{\rho_{\rm f}} } $
Work Step by Step
First, we need to find the buoyant force exerted on the cone. We know that this buoyant force is equal to the cone's weight since it is floating in equilibrium.
$$\sum F_{y,cone}=F_B-mg=ma_y=m(0)=0$$
Hence,
$$F_B=mg\tag1$$
And we know, from Archimedes’ principle, that the buoyant force is given by the weight of the displaced liquid.
$$F_B=m_{\rm f}g=\rho_{\rm f}V_{1}g$$
where $V_1$ is the volume of the displaced liquid which is the volume of the submerged part of the cone under the liquid surface as well.
So,
$$F_B=\rho_{\rm f}V_1g$$
Plugging into (1);
$$\rho_{\rm f}V_1\color{red}{\bf\not}g=m\color{red}{\bf\not}g $$
Thus,
$$\rho_{\rm f}V_1 =m $$
where $m=\rho_{\rm o}V$
$$\rho_{\rm f}V_1 =\rho_{\rm o}V\tag 2 $$
Now we need to find the volume of the submerged part $V_1$ and as we see in the figure below, the whole volume of the cone is given by
$$V=V_1+V_2$$
So,
$$V_1=V-V_2$$
where $V=\frac{1}{3}Al$ and $V_2=\frac{1}{3}A_2h$
where $A$ is the cross-sectional area of the cone's base.
$$V_1=\frac{1}{3}Al-\frac{1}{3}A_2h$$
$$V_1=\frac{1}{3}(\pi r^2)l-\frac{1}{3}(\pi r_2^2)h $$
$$V_1=\frac{1}{3} \pi \left[r^2 l- r_2^2 h\right]\tag 3$$
From the geometry of the figure below, we can see that
$$\tan\theta=\dfrac{r}{l}=\dfrac{r_2}{h}$$
Hence,
$$r_2=\dfrac{rh}{l}$$
Plugging into (3);
$$V_1=\frac{1}{3} \pi \left[r^2 l- \left(\dfrac{rh}{l}\right)^2 h\right] $$
$$V_1=\frac{1}{3} \pi \left[r^2 l- \dfrac{r^2h^3}{l^2} \right] $$
$$V_1=\frac{1}{3} \pi \left[ \dfrac{r^2 l^3-r^2h^3}{l^2} \right] $$
$$V_1=\frac{1}{3} \pi r^2\left[ \dfrac{l^3-h^3}{l^2} \right] $$
Plugging into (2);
$$\frac{1}{3} \pi r^2\left[ \dfrac{l^3-h^3}{l^2} \right]\rho_{\rm f} =\rho_{\rm o}V $$
Plug the volume of the whole cone;
$$\frac{1}{3} \pi r^2\left[ \dfrac{l^3-h^3}{l^2} \right]\rho_{\rm f} =\frac{1}{3} \pi r^2 l\rho_{\rm o} $$
$\frac{1}{3} \pi r^2$ cancels from both sides;
$$ \dfrac{l^3-h^3}{l^2} \rho_{\rm f} = l\rho_{\rm o} $$
$$ \dfrac{l^3-h^3}{l^3} =\dfrac{ \rho_{\rm o}}{\rho_{\rm f}} $$
$$ 1-\dfrac{h^3}{l^3} =\dfrac{ \rho_{\rm o}}{\rho_{\rm f}} $$
$$ \dfrac{h^3}{l^3} =1-\dfrac{ \rho_{\rm o}}{\rho_{\rm f}} $$
Therefore,
$$\boxed{ \dfrac{h }{l } =\sqrt[3]{1-\dfrac{ \rho_{\rm o}}{\rho_{\rm f}} }}$$
