The correct answer is C.

Probability of default = $\text{N}\left(\frac{\ln(F) - \ln(V) - \mu(T\text{-}t) + 0.5\sigma^2(T\text{-}t)}{\sigma\sqrt{T\text{-}t}}\right)$ = $\text{N}\left(\frac{\ln(80) - \ln(100) - 0.2(3) + 0.5(0.3^2)(3)}{0.3\sqrt{3}}\right)$ = $\text{N}(-1.3243) = 1 - \text{N}(1.3243) = 1 - 0.9073 = 0.0927 = 9.27\%$

$\text{LGD} = F \times \text{PD} - V e^{\mu(T-t)} \times \text{N}\left(\frac{\ln(F) - \ln(V) - \mu(T\text{-}t) - 0.5\sigma^2(T\text{-}t)}{\sigma\sqrt{T\text{-}t}}\right)$ = $80(0.0927) - 100 e^{0.2 \times 3} \times \text{N}\left(\frac{\ln(80) - \ln(100) - 0.2(3) - 0.5(0.3^2)(3)}{0.3\sqrt{3}}\right)$= `$7.416 - 182.2119 \times \text{N}(-1.8439)$ = $7.41`6 - 182.2119 \times (1 - 0.9673)$ = $7.416 - 182.2119 \times 0.0327$= `$7.416 - 6.0183$ = $\$1.45`77\text{m}$