Answer
$7.5\;\rm mA$
Work Step by Step
a) We know that
$$V_B=V_D+V_R$$
where $B$ is for battery, $D$ for diode, and $R$ for resistor.
So,
$$2=V_D+IR$$
$$2=V_D+180I $$
Solving for $I$;
$$I=\dfrac{2-V_D}{180}$$
Now we need to draw the graph of $I$ as a function of $V_D$, as we see below.
We have here two intersection points of our line, the first with $y$-axis at $(0\;\rm V,11.11\; mA)$ and the other with the original curve of figure 29-30.
As we see below, we draw a small part of the original graph in red just to show the intersection point with it.
This point is approximately at $0.65\;\rm V$.
Plugging into (1);
$$I=\dfrac{2-0.65}{180}=\color{red}{\bf 7.5}\;\rm mA$$