Current across a diode?
You are asked to find the operating point such that the small-signal resistance is 5 ohms, i.e.
$$\frac{dv_d}{di_d} = 5\,[\Omega]$$
You can obtain this by both sides of the given equation relating voltage and current, then rearranging as needed. The most straightforward way is to rearrange the equation above to a transconductance equation:
$$\begin{align}\frac{di_d}{dv_d} = 0.2\,[\Omega^{-1}] &amp;= \frac{d}{dv_d}\left[I_s \left(e^{\frac{v_d}{v_{th}}}-1\right)\right]\\
0.2\,[\Omega^{-1}] &amp;=  \frac{I_s}{v_{th}} \left(e^{\frac{v_d}{v_{th}}}\right)\\
0.2\,[\text{A}/\text{V}] &amp;= \frac{8\cdot10^{-12}\,[\text{A}]}{0.02\,[\text{V}]} \cdot e^{\frac{v_d}{0.02\,[\text{V}]}}\\
5\cdot10^8 &amp;= e^{\frac{v_d}{0.02\,[\text{V}]}}\\
\ln{5\cdot10^8}&amp;=\frac{v_d}{0.02\,[\text{V}]}\\
{0.02\,[\text{V}]}\cdot\ln{\left(5\cdot10^8\right)} &amp;= v_d 
\end{align}$$
There is now only one unknown on the right side, specifically \$v_d\$ (because \$I_s\$ and \$v_{th}\$ are provided); by substituting back into the original equation you can also find \$i_d\$.
The Wikipedia article on diode modeling discusses this further, along with a discussion of how one would model AC effects such as the junction capacitance. This sort of small-signal modelling is very useful in analog design, allowing you to analyze more complex devices such as transistors, and complex circuits such as op-amps.