1. Given data

• Zener voltage: $V_Z = 8\,\text{V}$
• Maximum power dissipation: $P_Z = 0.5\,\text{W}$

We need to find the protective resistance $R_p$.

2. Maximum safe current through zener diode

At the rated condition,

$P_Z = V_Z I_Z$

So the maximum zener current is

$I_{Z,\max} = \frac{P_Z}{V_Z} = \frac{0.5}{8} = 0.0625\,\text{A}$

Thus,

$I_{Z,\max} = 62.5\,\text{mA}$

3. Voltage across protective resistor

From the given circuit (potential divider arrangement), the maximum potential difference available across the zener branch is $20\,\text{V}$.

Since the zener maintains $8\,\text{V}$ across itself, the remaining voltage appears across $R_p$:

$V_{R_p} = 20 - 8 = 12\,\text{V}$

4. Calculate protective resistance

Using Ohm’s law,

$R_p = \frac{V_{R_p}}{I_{Z,\max}} = \frac{12}{0.0625}$

$R_p = 192\,\Omega$

5. Final answer

$\boxed{192\,\Omega}$

So the required integer answer is:

$\boxed{192}$