1. Given data

• Zener breakdown voltage: $V_Z = 10\,\text{V}$
• Power rating of zener: $P_Z = 2\,\text{W}$
• Input voltage fluctuates between $6\,\text{V}$ and $14\,\text{V}$
• We need series resistance $R_s$ for safe operation.

2. Maximum safe zener current

The zener diode must not dissipate more than its rated power.

$P_Z = V_Z I_{Z,\max}$

So,

$I_{Z,\max} = \frac{P_Z}{V_Z} = \frac{2}{10} = 0.2\,\text{A}$

3. Condition for worst-case current

The zener current is maximum when the supply voltage is maximum.

Here, maximum input voltage is:

$V_{in,\max} = 14\,\text{V}$

At this condition,

$I_Z = \frac{V_{in,\max} - V_Z}{R_s}$

For safe operation,

$I_Z \le I_{Z,\max} = 0.2\,\text{A}$

Thus,

$\frac{14 - 10}{R_s} \le 0.2$

$\frac{4}{R_s} \le 0.2$

$R_s \ge \frac{4}{0.2} = 20\,\Omega$

4. Final answer

For safe operation, the minimum required series resistance is:

$R_s = 20\,\Omega$

So the integer answer is:

$\boxed{20}$

5. Comparison with stored answer

Stored correct answer = $20$

This matches our derived answer.