Answer
$\approx 14.13\;\rm cm$
Work Step by Step
Let's assume that the boat is perfectly as shaped as a rectangular box, as we see in the figure below.
We can see that $L$ is the boat's length, $W$ is its width, $t$ is the thickness of its bottom, $t'$ is the thickness of the 4 walls, $h$ is the height of its walls, and $h'$ is the submerged part of it in the water.
To make this boat float in perfectly calm water means that the net vertical force exerted on the boat is zero.
Hence, the buoyant force must equal the net weight of the boat.
$$F_B=Mg\tag 1$$
We know, from Archimedes’ principle, that the buoyant force is equal to the weight of the displaced water due to the submerged part of the object.
So, $$F_B=m_{\rm displaced} g=\rho_{\rm water}V_{\rm submerged } g=\rho_{\rm water}Ah'g$$
where $A$ here is the cross-sectional area of the bottom of the boat and it is given by $A=LW$,
$$F_B=\rho_{\rm water}LWh'g$$
Plugging into (1);
$$\rho_{\rm water}LWh'\color{red}{\bf\not} g=M\color{red}{\bf\not} g $$
Hence,
$$h'=\dfrac{M}{\rho_{\rm water}LW}\tag 2$$
Now we need to find the total mass $M$ of the boat,
$\bullet$ First, the mass of the bottom,
$m_b=\rho V=\rho A t=\color{blue}{\rho Lwt}$
$\bullet$ second, the masses of the walls of side 1, where we have two walls of side 1.
$m_1=2\rho V_1=2\rho A_1 t'=\color{blue}{2\rho Lht'}$
$\bullet$ third, the masses of the walls of side 2, where we have two walls of side 2.
$m_2=2\rho V_2=2\rho A_2 t'=\color{blue}{2\rho Wht'}$
Therefore,
$$M=\rho Lwt+2\rho Lht'+2\rho Wht'$$
Plugging into (2);
$$h'=\dfrac{\rho (Lwt+2 Lht'+2 Wht')}{\rho_{\rm water}LW} $$
Plugging the known;
$$h'=\dfrac{(7900)([(10)(5)(0.02)]+[2 (10)(0.005)h]+[2(5)(0.005)h ])}{(1000)(10)(5)} $$
$$h'=\dfrac{(7900)(1+0.1h+0.05h )}{50000 } =0.158(0.15h+1)$$
$$h'=0.0237h+0.158$$
where $h'=h+t$
$$h+t=0.0237h+0.158$$
$$h-0.0237h=0.158-0.02$$
Thus,
$$h=0.1413\;\rm m\approx \color{red}{\bf 14.13}\;\rm cm$$
Therefore, the minimum height of the sides that makes this boat float in perfectly calm water is 14 cm.
