Required probability = P(second coin = fair / first coin shows head)
By conditional probability theorem, P(A / B) = $\frac{P(A \cap B)}{P(B)}$
$\therefore$ P(second coin = fair / first coin shows head) = $\frac{P(second coin = fair \cap first coin shows head)}{P(first coin shows head)}$
Now consider,
P(first coin shows head) = probability of getting a head on a fair coin + probability of getting a head on a biased coin
= $\frac{5}{10} * \frac{1}{2} + \frac{5}{10} * \frac{4}{5}$
= $\frac{13}{20}$
Now, P(second coin = fair $\cap$ first coin shows head)
If we observe closely, we can conclude that these probabilities are independent probabilities and for two independent events A and B, P(A $\cap$ B) = P(A) * P(B)
$\therefore$ P(second coin = fair $\cap$ first coin shows head)
= P(first coin shows head) * P(second coin = fair)
= $\frac{5}{10} * \frac{1}{2} * \frac{4}{9}$ +$\frac{5}{10} * \frac{4}{5} * \frac{5}{9}$
= $\frac{1}{3}$
$\therefore$ Required probability = $\frac{\frac{1}{3}}{\frac{13}{20}}$ = $\frac{20}{39}$
So, a) is correct.