How to solve using condition probability

No, they are not independent. He can claim Six when he get actually six or claim six when he get another number. Using Bayes' theorem:

$$P(\text{Six}|\text{Claimed Six})=\frac{\frac{1}{6}\frac{3}{4}}{\frac{1}{6}\frac{3}{4}+\frac{5}{6}\frac{1}{4}}=\frac{3}{3+5}=\frac{3}{8}$$

Your answer is not correct because of your small misundertanding of independence, since if ypu say that events $T$ and $D$ are independents, then that means: $$P(D|T)=P(D)$$ But at this time $D$ and $T$ are not independents because you dont have the outcome landed by the die but a stament wich is true with probability 3/4. So you need to use Bayes's Formula here as highlighted by @tommik.