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.