Size of Disk Block =1024 Byte
Disk Blocks address = 32 bit, but 48 bit integers are used for address
Therefore address size =48/8= 6 bytes9 (We always refer the address size )
No. of addresses per block 1024/6 = 170 addresses (approx)
Maximum File Size = 10 Direct + 1 Single Indirect + 1 Double Indirect + 1 Triple Indirect
= 10 + 170 + 170*170 + 170*170*170
≈ 2^22 Blocks
Since each block is of size 2^10 B
Maximum files size = 2^22 * 2^10 = 2^32 B
also, this is an approx answer