Answer
$5.22\;\rm cm$
Work Step by Step
We know that the can is floating upright in water when it is half full of water.
This means that the net force exerted on it is zero.
So,
$$F_B=mg$$
where $F_B$ is the buoyant force and $mg$ is the weight of the can plus the water inside it. So $m=m_c+m_w$ whreas $m_c$ is the can's mass and $m_w$ is the mass of the water inside the can,
$$F_B=(m_c+m_w)g\tag 1$$
We know, from Archemides' principle, that the buoyant force is given by the weight of the displaced water due to the submerged part of the object.
$$F_B=m_{\rm displaced}g=\rho_w V_{\rm submerged }g$$
Let's assume that the submerged length of the can is $h$, so
$$F_B= \rho_w Ah g$$
where $\rho_w$ is the water density.
Plugging into (1);
$$\rho_w Ah\color{red}{\bf\not} g=(m_c+m_w)\color{red}{\bf\not} g$$
$$\rho_w Ah =m_c+m_w$$
The author asks for the length of the can above the water level $h'$, and we know that the full length of the can is $L=h+h'$, so the length of the submerged part in the water is given by $h=L-h'$
$$\rho_w A(L-h') =m_c+m_w$$
Solving for $h'$;
$$h'=L-\dfrac{m_c+m_w}{\rho_w A}\tag 2$$
Now we need to find the mass of the water inside the can;
$$m_w=\rho_w V$$
where $V$ is the half the can's volume, so $V=\frac{1}{2}AL$. Hence,
$$m_w=\frac{1}{2}AL \rho_w $$
Plugging into (2):
$$h'=L-\dfrac{m_c+\frac{1}{2}AL \rho_w}{\rho_w A} $$
Recall that the cross-sectional area of the can is given by $A=\pi r^2$
$$h'=L-\dfrac{m_c+\frac{1}{2}\pi r^2L \rho_w}{\rho_w \pi r^2} $$
Before plugging the known, we need to find the can's length since we know its cross-sectional area and its volume.
So, $V=AL$, and hence,
$$L=V/A=\dfrac{355\times 10^{-3}\times 10^{-3}}{\pi r^2}=\dfrac{355\times 10^{-6}}{\pi \left(\dfrac{6.2}{2}\times 10^{-2}\right)^2}$$
$$L=0.1176\;\rm m$$
Plugging the known;
$$h'=(0.1176)-\dfrac{[20\times 10^{-3}]+\frac{1}{2}\pi \left(\dfrac{6.2}{2}\times 10^{-2}\right)^2(0.1176) (1000)}{\pi \left(\dfrac{6.2}{2}\times 10^{-2}\right)^2(1000)} $$
$$h'=0.0522\;\rm m\approx \color{red}{\bf 5.22}\;\rm cm$$
