1. Circuit idea

   For a zener regulator,

   • Input voltage: $V_i = 10\,\text{V}$
   • Zener voltage: $V_Z = 8\,\text{V}$
   • Series resistance: $R = 100\,\Omega$
   • Maximum allowed zener current: $I_{ZM} = 10\,\text{mA}$

   Since the zener regulates, the load voltage is approximately $V_L = V_Z = 8\,\text{V}.$

2. Current through series resistor

   The current supplied through the series resistor is $I = \frac{V_i - V_Z}{R} = \frac{10 - 8}{100} = \frac{2}{100} = 0.02\,\text{A} = 20\,\text{mA}.$

   This current splits as $I = I_Z + I_L.$

3. Condition for maximum load resistance $R_{L,\max}$

   Load current is $I_L = \frac{V_Z}{R_L}.$

   For maximum $R_L$, the load current is minimum, so zener current becomes maximum.

   To keep zener safe, $I_Z \le I_{ZM} = 10\,\text{mA}.$

   At the limiting condition for maximum $R_L$, $I_Z = 10\,\text{mA}.$

   Hence, $I_L = I - I_Z = 20 - 10 = 10\,\text{mA}.$

   Therefore, $R_{L,\max} = \frac{V_Z}{I_L} = \frac{8}{10\times 10^{-3}} = 800\,\Omega.$

4. Condition for minimum load resistance $R_{L,\min}$

   For minimum $R_L$, load current is maximum, so zener current is minimum.

   For regulation, the minimum zener current can be taken as $I_Z = 0.$

   Then, $I_L = I = 20\,\text{mA}.$

   So, $R_{L,\min} = \frac{V_Z}{I_L} = \frac{8}{20\times 10^{-3}} = 400\,\Omega.$

5. Required ratio

   $\frac{R_{L,\max}}{R_{L,\min}} = \frac{800}{400} = 2.$

Final Answer: $\boxed{2}$