You can solve this by the application of Bayes' theorem.
You are given the probability of choosing coins $A$ and $B$:
$$P(A) = P(B) = \frac{1}{2}$$
You are also given the probability of getting a head, given that you have chosen coin $A$ or $B$:
$$P(H|A) = \frac{6}{10} \quad P(H|B) = \frac{1}{2}$$
Therefore you can compute the probability of getting 10 heads with either coin.
$$P(10 H|A) = \left(\frac{6}{10}\right)^{10} \quad P(10 H|B) = \left(\frac{1}{2}\right)^{10}$$
You are interested in the probability that coin A was used, given that 10 heads were observed, i.e. $P(A|10H)$. Using Bayes' theorem:
$$P(A|10H) = \frac{P(10H|A)P(A)}{P(10H)}$$
The marginal probability $P(10H)$ is the weighted sum of probabilies of getting 10 Heads across the two coins: $$P(10H)=P(10H|A)P(A)+P(10H|B)P(B)$$
Therefore:
$$\frac{P(10H|A)P(A)}{P(10H|A)P(A)+P(10H|B)P(B)}$$
And since in this example $P(A) = P(B)$
$$P(A|10H)=\frac{P(10H|A)}{P(10H|A)+P(10H|B)}= \frac{\left(\frac{6}{10}\right)^{10}}{\left(\frac{6}{10}\right)^{10} + \left(\frac{1}{2}\right)^{10}} \approx 0.861$$