V-I characteristics of a silicon diode is as shown.

 

The ratio of resistance of diode at ${\mathrm{I}}_{\mathrm{D}}=15 \mathrm{mA} \mathrm{and} {\mathrm{V}}_{\mathrm{D}}=-10\mathrm{V}, \mathrm{is}$

From the graph, given in question at I = 20mA, V is 0.8V and at I = 10 mA, V = 0.7 V

$$\begin{gathered} \therefore {\mathrm{r}}_{\mathrm{fb}}=\frac{∆\mathrm{V}}{∆\mathrm{I}}=\frac{0.8-0.7}{20-10}=\frac{0.1 \mathrm{V}}{10 \mathrm{mA}} \\ =\frac{0.1 \mathrm{V}}{10\times {10}^{-3} \mathrm{A}}=10 \mathrm{\Omega } \\ \mathrm{Also}, \mathrm{at} \mathrm{V}=-10\mathrm{V}, \mathrm{I}=-1 \mathrm{\mu A} \\ {\mathrm{r}}_{\mathrm{rb}}=\frac{10 \mathrm{V}}{1 \mathrm{\mu A}}=\frac{10 \mathrm{V}}{1\times {10}^{-6} \mathrm{A}}=1\times {10}^7 \mathrm{\Omega } \\ \therefore \mathrm{Ratio}=\frac{{\mathrm{r}}_{\mathrm{fb}}}{{\mathrm{r}}_{\mathrm{rb}}}=\frac{10}{{10}^7}={10}^{-6} \end{gathered}$$