8 Direct Disk Blocks
Single Indirect
Single, double indirect disk block
DBA: 16 bits= 2B
DB size = $2^12$B = 4KB
Maximum File size will be
no. of addresses per block = 4KB/2B = 2K
Max file size, when file is stored on single, double indirect disk block
= $2^{11}*2^{11}$ * 4KB = $2^{22} * 2^{12}$ B = $2^{34}$B = 16GB