To answer your question, you can't use the Edgeworth box in this kind of setup because it is used for an economy where we have two goods and two agents. However, there exists an alternative graphical representation to solve such problems.
To see how we arrive at the above graph, first, plot the utility function for individual 1, i.e., $U_1(x_1)$. Next, we plot the utility function for individual 2, but to get the picture above, we horizontally flip the graph of $U_2(x_2)$. This sort of diagram helps us find the set of Pareto efficient allocations. In the picture above, notice that for $x_1\in[2,12]$ and $x_2\in[8,18]$, an increase in the utility of individual 1 leads to a fall in the utility of individual 2 and vice versa.
Therefore, the green highlighted portion in the above diagram is the set of Pareto efficient locations, it can be written as $$\boxed{\mathcal{P}=\{(x_1,x_2)\in \mathbb{R}_+^2 \mid x_1\in [2,12] \land x_2=20-x_1\}}$$
Formally, the set of Pareto Efficient allocations is the set of all those allocations which maximize the utility of both individuals subject to the feasibility constraint.
In what follows, I use $\mathcal{F}$ to denote the set of all feasible allocations.
A set $\mathcal{P} \subseteq \mathcal{F}=\{(x_1,x_2)\in \mathbb{R}_+^2 \mid x_1+x_2=20\}$ is called the set of Pareto Efficient allocations if for all $(x_1^*,x_2^*)\in\mathcal{P}$ the following holds:
Given $x_2^*$, $x_1^*$ is a solution to the problem $-$ $$\begin{align}\max_{x_1\geq 0} \quad & U_1=-|x_1-2| \\ \textrm{s.t.} \quad & x_1+x_2^*=20\end{align}$$
Given $x_1^*$, $x_2^*$ is a solution to the problem $-$
$$\begin{align} \max_{x_2\geq0} \quad & U_2=-|x_2-8| \\ \textrm{s.t.} \quad & x_1^*+x_2=20\end{align}$$
You can use the above definition to solve for Pareto Efficiency analytically.
For part b) we need to solve the following:
$$\begin{align}
\max_{x_1,x_2\geq 0} \quad & \min(U_1(x_1),U_2(x_2)) \\
\textrm{s.t.} \quad & x_1+x_2=20 \\ \\
\max_{x_1,x_2\geq 0} \quad & \min(-|x_1-2|,-|x_2-8|) \\
\textrm{s.t.} \quad & x_1+x_2=20 \\ \\
\max_{0 \leq x_1 \leq 20} \quad & \min(-|x_1-2|, -|12-x_1|) & [\text{using substitution}]
\end{align}$$
The above problem is solved at $x_1=7$.
Thus, the professor chooses $\boxed{(x_1,x_2)=(7,13)}$ in order to maximise rawlsian social welfare function for the above setup.