The author, there is no need to go to the 256-bit address. 160 bits is enough to ensure the security of the address.
The problem is completely contrived.
In order to get a collision with a probability of 1.77636E-15 (and this probability is negligible, there is a higher chance of winning the lottery) there must be a blockchain in which 2 ^ 56 addresses will be used, each address being 20 bytes (160 bit). In total, the blockchain with such an amount of addresses will occupy not less than 2 ^ 56 * 20 = 1441151 TB (this is the lowest estimate), by the way, at the moment, the blockchain is only 126 GB.
When you read the thread did you see this post:
There are only 2,100,000,000,000,000 satoshis.
Therefore the maximum number of Bitcoins addresses that can contain value at any one time is only about 251 and that assumes only one satoshi per address.
At the time the block chain contains 255 transactions it would be somewhere between a petabyte and an exabyte in size. So a full node would need a multi petabyte drive to store the entire block chain.
Your estimate for block chain size is interesting. I will ponder that. Thanks.