Answer
See the detailed answer below.
Work Step by Step
First of all, we need to recall the induced potential difference through an inductor which is given by
$$\Delta V_L=-L\dfrac{dI}{dt}$$
$$\Delta V_L=-L\dfrac{\Delta I}{\Delta t}$$
Plug the known;
$$\Delta V_L=-(10\times 10^{-3})\dfrac{\Delta I}{\Delta t}\tag 1$$
We have, from the given graph, 4 stages. We will use (1) to find the potential difference in each stage.
$\bullet$ For the first stage from $t=0$ s to $t=2$ ms.
$$(\Delta V_L)_1=-(10\times 10^{-3})\dfrac{(2-0)}{(2-0)\times 10^{-3}}=\bf -10\;\rm V $$
$\bullet\bullet$ For the second stage from $t=2$ ms to $t=3$ ms.
$$(\Delta V_L)_2=-(10\times 10^{-3})\dfrac{(2-2)}{(3-2)\times 10^{-3}}=\bf 0\;\rm V $$
$\bullet\bullet\bullet$ For the third stage from $t=3$ ms to $t=5$ ms.
$$(\Delta V_L)_2=-(10\times 10^{-3})\dfrac{(-2-2)}{(5-3)\times 10^{-3}}=\bf 20\;\rm V $$
$\bullet\bullet\bullet\bullet$ For the fourth stage from $t=5$ ms to $t=6$ ms.
$$(\Delta V_L)_3=-(10\times 10^{-3})\dfrac{(-2-[-2])}{(6-5)\times 10^{-3}}=\bf 0\;\rm V $$
Now we can draw the graph of the potential difference through these four stages.
See the graph below.