Find the first set of 5 digits in a text

To match the first 5-digit number you may use

^.*?\K(?<!\d)\d{5}(?!\d)

See the regex demo. As you want to remove the match, simply keep the Replace With field blank. The ^ matches the start of a line, .*? matches any 0+ chars other than line break chars, as few as possible, and \K operator drops the text matched so far. Then, (?<!\d)\d{5}(?!\d) matches 5 digits not enclosed with other digits.

Another variation includes a capturing group/backreference:

Find What:      ^(.*?)(?<!\d)\d{5}(?!\d)
Replace With: $1

See this regex demo.

Here, instead of dropping the found text before the number, (.*?) is captured into Group 1 and $1 in the replacement pattern puts it back.

To find the first 5 digits in the text, you could also match not a digit \D* or 1-4 digits followed by matching 5 digits:

^(?=.*\b\d{5}\b)(?:\D*|\d{1,4})*\K\d{5}(?!\d)

• ^ Start of string
• (?=.*\b\d{5}\b) Assert that there are 5 consecutive digits between word boundaries
• (?:\D*|\d{1,4})* Repeat matching 0+ times not a digit or 1-4 digits
• \K\d{5} Forget what was matched, then match 5 digits
• (?!\d) Assert what followed is not a digit

Regex demo

I would've use

(^(?:(?!\d{5}).)+)(\d{5})(?!\d)

It finds fragment from beginning of the string till end of first 5-digit number, but in case of replacement you can use $1 or $2 to substitute corresponding part. For example replacement $1<$2> will surround number by < and >.