Answer
See the answer below.
Work Step by Step
The net force exerted on the boat and its component is always zero.
$$\sum F_y=F_B-m_bg=0$$
$$F_B=m_bg$$
where $m_b$ is the mass of the boat and its component which is given by $m_b=M+Nm$ where $M$ is the boat's mass without the boxes, $N$ is the number of the boxes, and $m$ is the mass of one box.
$$F_B=(M+Nm)g\tag1$$
According to Archimedes’ principle, the buoyant force is given by
$$F_B= m_{\rm displaced}g=\rho V g$$
where $\rho$ is the liquid density and $V $ is the submerged part of the boat's volume.
This volume is given by $V=Ah$ where $A$ is the cross-sectional area of the boat and $h$ is the depth of the submerged part of the boat.
Hence,
$$F_B= Ah\rho g$$
Plugging into (1);
$$Ah\rho \color{red}{\bf\not} g=(M+Nm)\color{red}{\bf\not} g$$
Thus,
$$Nm=A\rho h-M$$
Now we have a straight-line equation of $y=mx+a$;
where $y=Nm$, $m=A\rho$ which is the slope of the line, and $a=-M$ which is the $y$-intercept.
So to find the boat's mass, we need to draw this straight line and find its $y$-intercept.
\begin{array}{|c|c|c|}
\hline
h\;({\rm m})& Nm\;{\rm (kg)} \\
\hline
0.029 & 0.050 \\
\hline
0.05 & 0.100 \\
\hline
0.066 & 0.150 \\
\hline
0.068& 0.200 \\
\hline
\end{array}
Plot these data into any software calculator.
Then find the slope that will give us the density of the liquid.
See the graph below, the best-fit line there, and its slope.
From the software calculator and the figure below, it is obvious that
$$M=-a=-(-29.03\times 10^{-3})=0.02903\;\rm kg$$
Hence,
$$M=\color{red}{\bf 29}\;\rm g$$
And the slope is
$${\rm Slope}=2.67=A\rho$$
So,
$$\rho=\dfrac{2.67}{A}=\dfrac{2.67}{25\times 10^{-4}} $$
$$\rho=\color{red}{\bf 1068}\;\rm kg/m^2$$
