Interpret the regulator circuit

Since the circuit is stated to give an 8 V regulated dc output using two identical zener diodes, the most natural configuration is that the two identical zeners are connected in series in reverse breakdown across the output.

Hence, if each zener has breakdown voltage $V_Z$, then $2V_Z = 8\text{ V}$ so $V_Z = 4\text{ V}.$

Input and series resistor drop

The input is $12\text{ V}$ and the regulated output is $8\text{ V}$, so the remaining voltage is dropped across the series resistor $R$: $V_R = 12 - 8 = 4\text{ V}.$

Current through the zener branch

From the given standard regulator setup (series resistor of $100\,\Omega$ in the shown circuit), the current through the resistor and hence through the zener string is $I = \frac{V_R}{R} = \frac{4}{100} = 0.04\text{ A} = 40\text{ mA}.$

Power dissipated in each diode

Each identical zener diode has voltage drop $4\text{ V}$ and carries current $40\text{ mA}$.

Therefore, power dissipated in each diode is $P = V_Z I = 4 \times 0.04 = 0.16\text{ W} = 160\text{ mW}.$

Comparison with stored answer

My derived value is $\boxed{160}$ mW for each diode.

This does not match the stored correct answer $40$.

Why the stored answer may be incorrect

For the answer to be $40\text{ mW}$, each diode would need to dissipate $P=40\text{ mW}=0.04\text{ W}.$ Since each diode has $4\text{ V}$ across it, the current would then have to be $I=\frac{0.04}{4}=0.01\text{ A}=10\text{ mA},$ which would require the series resistor to be $R=\frac{4}{0.01}=400\,\Omega.$

So unless the resistor in the figure is actually $400\,\Omega$, the stored answer $40$ is inconsistent with the usual regulator calculation.

Based on the standard reading of the circuit, the correct answer should be $160\text{ mW}$ per diode.