Voltage divider in series circuit Steps:

• PD across LDR: $V_{LDR} = 0.6 \times \frac{R_{LDR}}{R_{thermistor} + R_{LDR}}$.
• Compute for each option to find where $V_{LDR} = 0.2$ V.
• Option A: $R_{thermistor} = 600~\Omega$, $R_{LDR} = 500~\Omega$, total = 1100 , V.
• Option B: total = 2600 $\Omega$, $V_{LDR} \approx 0.46$ V.
• Option C: total = 600 $\Omega$, $V_{LDR} = 0.5$ V.
• Option D: total = 2100 $\Omega$, $V_{LDR} \approx 0.57$ V. Not enough information (no diagram to confirm configuration; calculations do not yield exactly 0.2 V for any option).

Final answer: Not enough information.