A telephone number contains 10 digit, including a 3-digit

If you delete those words, then it's not clear what the question is asking. Do you want the probability he gets the phone number right precisely on his second attempt? Or in one of his first two attempts? Here we have 25 possibilities, one of which is right. If he'll pick two of the numbers, he'll have a 2/25 chance to pick the right one. Mathematically, the problem is essentially identical to a different question you asked about yesterday, which was something like this - if you have 25 employees, including Bob, and randomly pick a President and Vice-President, what is the probability Bob will be President or VP? The answer is 2/25.

Some solutions above produce the answer 49/625, which is not correct, unless he might pick the same phone number twice. You can see why that method is wrong just by imagining there are only two possible phone numbers. Then in two attempts Bob should have a 100% chance of getting the number right, but if you use the method that produces the answer 49/625, you would get an answer less than 100% to the two phone number question.

I have no idea why, in the original question, they appear to be asking for an estimate, nor why the fractions in the answer choices are not reduced.