Let $HH$ be the event that we get $2$ heads in a row. Let $H_3$ be the event the third toss is a head. We want $\Pr(H_3|HH)$. Maybe we start from something a little simpler than the Bayes' Theorem that you used, essentially the definition of conditional probability: $\Pr(H_3|HH)=\frac{\Pr(H_3 \cap HH)}{\Pr(HH)}.$
We calculate the two probabilities on the right. For $\Pr(HH)$, note that two heads in a row happens with probability $(4/5)^2$ if we use the funny coin, and with probability $(1/2)^2$ if the coin is the ordinary coin. It follows that $\Pr(HH)=\frac{1}{2}\left(\frac{4}{5}\right)^2+\frac{1}{2}\left(\frac{1}{2}\right)^2.$
A similar calculation gives us the probability of $HH$, followed by $H_3$: $\Pr(HH\cap H_3)=\frac{1}{2}\left(\frac{4}{5}\right)^3+\frac{1}{2}\left(\frac{1}{2}\right)^3.$ Divide.