nBit to leading zeros in Block Hash
The bits value you have written is in decimal, in hex it is 0x181b7b74

We split this into the exponent and mantissa as shown in this answer: https://bitcoin.stackexchange.com/a/61951/51948

0x18 in decimal is 24, so we calculate the target to be:
0x1b7b74000000000000000000000000000000000000000000

Now as the target takes up 24 bytes, out of the total 32 bytes for a block hash, the remaining 8 bytes must be 0's in order for the hash to be less than the target  (as well as the rest of the number being less than, of course). Usually leading 0's are omitted for readibility, but 8 zero bytes is written as 16 zero characters in hexadecimal, which is why the hash must have at least 16 leading zeros - which it does.