Answer: 0.03

B is correct. The calculation is: $$\text{put} = Ke^{-rT}N(-d_2) - SN(-d_1)$$ Given: - K = 20 - r = 4.25% = 0.0425 - T = 0.5 - S = 25 - N(-d_2) = 1 - 0.9651 = 0.0349 - N(-d_1) = 1 - 0.9737 = 0.0263 Step-by-step calculation: 1. $Ke^{-rT} = 20 \times e^{-0.0425 \times 0.5} = 20 \times e^{-0.02125} \approx 20 \times 0.97896 = 19.5792$ 2. $Ke^{-rT}N(-d_2) = 19.5792 \times 0.0349 \approx 0.683$ 3. $SN(-d_1) = 25 \times 0.0263 = 0.6575$ 4. $\text{put} = 0.683 - 0.6575 = 0.0255 \approx 0.03$ The put option value is approximately 0.03.

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