Where's my mistake? This is partially an electrical engineering problem, but the actual issue apparently lays in my maths.
I have the equation $\frac{V_T}{I_0} \cdot e^{-V_D\div V_T} = 100$
$V_T$ an $I_0$ are given and I am solving for $V_D$.
These are my steps:
Divide both sides by $\frac{V_T}{I_0}$: 
$$e^{-V_D\div V_T} = 100 \div \frac{V_T}{I_0}$$
Simplify: 
$$e^{-V_D\div V_T} = \frac{100\cdot I_0}{V_T}$$
Find the natural log of both sides: 
$$-V_D\div V_T = \ln(\frac{100\cdot I_0}{V_T})$$
Multiply by $V_T$: 
$$-V_D = V_T \cdot \ln(\frac{100\cdot I_0}{V_T})$$
Multiply by $-1$:
$$V_D = -V_T \cdot \ln(\frac{100\cdot I_0}{V_T})$$
Now if I substitute in the numbers; $V_T \approx 0.026$, $I_0 \approx 8 \cdot 10^{-14}$
$$V_D = -0.026 \cdot \ln(\frac{8 \cdot 10^{-12}}{0.026})$$
Simplify a little more:
$$ V_D = \frac{26}{1000} \cdot -\ln(\frac{4}{13} \cdot 10^{-9})$$
and you finally end up with $V_D \approx 0.569449942$
There is an extra step to the problem as well: the question calls not for $V_D$, but for $V_I$ which is a source voltage and that can basically be determined by solving this:
$$V_I \cdot \frac{1}{40} = V_D$$
I.e. $V_I = 40V_D$; which makes $V_I \approx 22.77799768$.
However, this is off by quite a bit (the answer is apparently $1.5742888791$).
Official Solution:

We find $V_D$ to be $\approx$ 0.57.

(Woo! I did get that portion right.)

Since we know that $\frac{V_D}{i_D}$ is 100, we find $i_D$ to be 0.00026.

Background: $\frac{V_D}{i_D}$ is the resistance that I was originally solving. $V_D$ was the voltage across the element, and $i_D$ was the current through the element.
However, either I'm making some stupid mistake but if $i_D = \frac{V_D}{100}$ then how did they get 0.00026?
Completing the rest of the solution's method (quite convoluted in comparison to mine, checked mine though another method; $V_I$ does in fact equal $40V_D$), with the correct value, 0.0057, I arrived at exactly the same final value as before.
Would it be fair to say that it is likely that my logic is correct?
 A: You are confusing resistance with incremental resistance, I think. The incremental resistance only matters for small signal analysis. The problem is to set the operating point so that the incremental resistance will be the required value. This involves computing $V_D, I_D$. However, you cannot use the incremental resistance to compute $I_D$ in terms of $V_D$.
Also, you are forgetting to account for the $3.9k\Omega$ series resistance.
You correctly computed $V_D = 0.57 V$ required so the incremental resistance is $100 \Omega$.
However, the current through the diode at $V_D$ is given by the diode equation, which you haven't included above. The diode equation is $I_D = I_0 (e^{V_D/V_T}-1)$. Plugging in your numbers gives $I_D = 260 \mu A$ (basically $\frac{V_T}{100}$, since $\frac{V_D}{V_T}$ is large).
The question was to figure the input voltage $V_I$ that will set the diode operating point at $V_D =0.57 V, I_D=260 \mu A$. In the first instance, there is a series resistance of $R_S = 3.9 k\Omega$, so to figure the required $V_I$ you need
$$ V_I = R_S I_D+V_D$$
Plugging in the numbers gives $V_I = 1.58V$ or so.
