Answer
a) See the graph below.
b) $x=3\;\rm m$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric potential energy is given by
$$U_e=qV$$
So we can use the main points in the given graph to draw a $U_e-x$ graph.
At $(x=0{\;\rm m},V=0{\;\rm V})$, $U_e=qV=(0.1)(0)=\bf 0\;\rm J$
At $(x=1{\;\rm m},V=10{\;\rm V})$, $U_e=qV=(0.1)(10)=\bf 1\;\rm J$
At $(x=2{\;\rm m},V= 0{\;\rm V})$, $U_e=qV=(0.1)(0)=\bf 0\;\rm J$
At $(x=3{\;\rm m},V= 20{\;\rm V})$, $U_e=qV=(0.1)(20)=\bf 2\;\rm J$
Now we have 4 dots, as shown below.
See the graph below.
$$\color{blue}{\bf [b]}$$
To find the turning point, we need to find the electric potential energy at this point,
Since the energy is conserved,
$$E_i=E_f$$
$$K_i+U_{ei}=K_f+U_{ef}$$
at the turning point $v=0$, so $K_f=0$
$$K_i+U_{ei}=0+U_{ef}$$
Hence,
$$U_{ef}=K_i+U_{ei}$$
At $x=1$, $K_i=1\;\rm J$, and $U_{e}=1\;\rm J$; so that
$$U_{ef}=1+1=\bf 2\;\rm J$$
From our graph, the electric potential energy equals 2 J at the point of
$$x=\color{red}{\bf 3}\;\rm m$$
