Define the Events:

Let $A$ be the event that an unbiased coin is drawn.

Let $B$ be the event that the coin with heads on both sides is drawn.

Let $H$ be the event that a head turns up when the coin is tossed.

State the Given Probabilities:

$P(A) = \frac{19}{20}$ (since there are 19 unbiased coins out of 20 total coins)

$P(B) = \frac{1}{20}$ (since there is 1 coin with two heads out of 20 total coins)

$P(H|A) = \frac{1}{2}$ (probability of getting a head given an unbiased coin is drawn)

$P(H|B) = 1$ (probability of getting a head given the coin with two heads is drawn)

Use Bayes' Theorem:

We want to find $P(A|H)$, which is the probability that the drawn coin was unbiased given that a head turned up. Bayes' Theorem states:

$$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)}$$

Calculate $P(H)$:

We can find $P(H)$ using the law of total probability:

$$P(H) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B)$$

$$P(H) = \left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right) + (1) \cdot \left(\frac{1}{20}\right) = \frac{19}{40} + \frac{1}{20} = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$$

Apply Bayes' Theorem to find $P(A|H)$:

$$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right)}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}$$

Determine $m$ and $n$:

Since $P(A|H) = \frac{m}{n}$ and $\gcd(m, n) = 1$, we have $m = 19$ and $n = 21$.

Calculate $n^2 - m^2$:

$$n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = (40)(2) = 80$$

Therefore, $n^2 - m^2 = 80$.

So, the correct answer is:

Option B: 80