Approximating a zener diode with an independent voltage source in series with a resistor
There must be a typographical error somewhere.  You are correct to doubt 10 Ohms.
Since Zeners become low resistance like a voltage source, you might simulate it as charging a battery with series resistance.  Of course, it would not generate power.  This is useful to remember because all real Zeners, batteries and voltage sources have a series resistance.   ( so technically a voltage sink)


Perhaps if the incremental slope was 0.1A/V then the Zener would have a knee resistance of Zzt = 10 Ω   ( the Zener impedance at a standard test current )

This is a more common value in discrete Zeners (than 1 Ohm),  as shown in a typical Zener datasheet.




It is customary to refer to forward current as a diode and reverse voltage breakdown as the Zener mode and thus often shown as a negative voltage.
... thus given RIN = 1 kΩ, RL = 2 kΩ, VI = 9.5 V, vi = 50 mV


If the Zener was Zzt=10 @20mA and the Thevenin equiv cct is 9.5V *2/3 @ R=1k//2k or Vs=  6.333V @ 0.666kΩ  thus the drop voltage is 1.333V and Zzt only raises Rs to 0.676 the DC current is 1.333V/0.676kΩ = 1.97mA ~ 2mA  

We could correct minor errors in actual load current vs rated current, but maybe next time.

Bonus

It is also useful to know that LED's rated for 20mA like the Zeners above have an incremental resistance of typically 15 Ohms but often more than twice the voltage tolerance due to a wide variation in this Zzt equivalent value.  This resistance value is inverse to power ratings and this series bulk R is the fundamental reason for Vf tolerances due to process controls.