What does this probability tree tells about the problem? I have this image of probability tree but I am unsure of what problem it solves.

From the given data I am assuming that it is telling that there's a six-sided die and it is thrown three times to get a desired number e.g. '6'.
So,
$=\Large\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$
Would this be the probability needed for getting '6' on a third trial? Otherwise, what does it represent? Assuming a die is involved and we are finding probability of getting a number.
 A: The "success" labels on the tree suggest that the problem may be to calculate the probability that you get at least one of your desired outcomes in the three tries.
If that's the right interpretation then you fail with probability
$(5/6)^3 = 125/216$ and succeed with the complementary probability $91/216$.
The path outlined in green is the case in which you succeed  on the third trial. You have correctly computed that probability.
Note that you need not be looking for the same number each time. Success might mean seeing $1$ on the first roll or $2$ on the second or $3$ on the third.
Edit to respond to this comment:

Just a bit of confusion. If trials are  $T1$, $T2$ and $T3$. I noticed that
probability of $P(T2|T1)=P(T2)$. So, events are both independent and
conditional?

I think that's not quite right.
"Both independent and conditional" doesn't make sense.
You have to think carefully about the sample space. If it's the full tree of possibilities with three dice rolls looking for a $1$ (say) then there are $2^3 = 8$ possible states, corresponding to success and failure on each roll. Then the probability of success on any particular roll is independent of what happens on any other. No roll gives information about another. So
$$
P(\text{success on roll } 2 \, | \, \text{success on roll } 1)= P(\text{success on roll } 2).
$$
But the picture suggests that you stop the experiment as soon as you see one success. Then the sample space has $4$ elements, corresponding to the $4$ leaves of that tree. If roll $1$ succeeds you don't roll a second time, so
$$
P(\text{success on roll } 2 \, | \, \text{success on roll }  1)
$$
doesn't make sense. You could set its probability to $0$. The probability of the subset of the sample space that corresponds to the existence of a second roll is not independent of what happens on the first roll. It's $5/6$ before that roll and either $1$ or $0$ after that roll. So
$$
P(\text{success on roll } 2 \, | \, \text{failure on roll }  1) = 1/6
$$
while
$$
P(\text{success on roll } 2)= \frac{5}{6}\frac{1}{6}.
$$
A: At each stem, the number represents the probability of that event happening. So $$\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$$ represents the probability of getting your desired number $6$ at your third throw.
