We can partition the events as follows:

Let's compute each term.
Winning at round 1
To win with the first die, Tammy must roll a 3. This happens with probability 1/6. So, we have

Winning at round 2
To win with the second die, Tammy must not roll a 3 with the first roll (probability 5/6), and then roll a 3 with the second roll (probability 1/6). So, we have

Winning at round 3
To win with the third die, Tammy must not roll a 3 with the first two rolls (probability 5/6), and then roll a 3 with the third roll (probability 1/6). So, we have

Winning at round 4
To win with the fourth die, Tammy must not roll a 3 with the first three rolls (probability 5/6), and then roll a 3 with the fourth roll (probability 1/6). So, we have

So, we have

Note
You can do this exercise more quickly by observing

And you lose if you never roll a 3 in 4 rolls (each failure has a probability of 5/6). In fact, you have
