The distance to default (DD) in the Merton model measures how many standard deviations a firm's asset value is away from the default threshold (the face value of its debt). A higher DD indicates a lower probability of default.

The standard distance to default calculation is: $DD = \frac{\ln(V/K) + (\mu - \frac{\sigma^2}{2})T}{\sigma \sqrt{T}}$

Let's calculate the DD for each company:

Company A:

• $V = 110$, $K = 30$, $\mu = 0.25$, $\sigma = 0.18$, $T = 1$
• $\ln(V/K) = \ln(110 / 30) = \ln(3.6667) \approx 1.2993$
• $(\mu - \frac{\sigma^2}{2}) = 0.25 - \frac{0.18^2}{2} = 0.25 - 0.0162 = 0.2338$
• $DD_A = \frac{1.2993 + 0.2338}{0.18} = \frac{1.5331}{0.18} \approx 8.517$

Company B:

• $V = 180$, $K = 120$, $\mu = 0.19$, $\sigma = 0.12$, $T = 1$
• $\ln(V/K) = \ln(180 / 120) = \ln(1.5) \approx 0.4055$
• $(\mu - \frac{\sigma^2}{2}) = 0.19 - \frac{0.12^2}{2} = 0.19 - 0.0072 = 0.1828$
• $DD_B = \frac{0.4055 + 0.1828}{0.12} = \frac{0.5883}{0.12} \approx 4.902$

Company C:

• $V = 300$, $K = 190$, $\mu = 0.30$, $\sigma = 0.09$, $T = 1$
• $\ln(V/K) = \ln(300 / 190) = \ln(1.5789) \approx 0.4568$
• $(\mu - \frac{\sigma^2}{2}) = 0.30 - \frac{0.09^2}{2} = 0.30 - 0.00405 = 0.29595$
• $DD_C = \frac{0.4568 + 0.29595}{0.09} = \frac{0.75275}{0.09} \approx 8.364$

Note: Alternatively, using a simplified approximation formula $DD \approx \frac{\ln(V/K)}{\sigma \sqrt{T}}$, the result yields $A \approx 7.21$, $C \approx 5.07$, $B \approx 3.38$. The ranking remains the same.

Comparing the values: $DD_A (8.517) > DD_C (8.364) > DD_B (4.902)$. Since a higher DD indicates a lower default likelihood, the ranking from least likely to default to most likely is A, C, B.