A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is

1. Block size $= 128\text{ Bytes}$.

2. Address size $= 8\text{ Bytes}$.

3. Number of addresses per disk block:

   $$N = \frac{128}{8} = 16$$

4. Calculate the maximum number of blocks addressable by a file descriptor:

   • Direct blocks:$8\text{ blocks}$

   • Single indirect blocks:$1 \times 16 = 16\text{ blocks}$

   • Double indirect blocks:$1 \times 16 \times 16 = 256\text{ blocks}$

5. Total number of blocks $= 8 + 16 + 256 = 280\text{ blocks}$.

6. Maximum possible file size:

   $$\text{Size} = 280 \times 128\text{ Bytes} = 35840\text{ Bytes} = \frac{35840}{1024}\text{ KBytes} = 35\text{ KBytes}$$

Hence, (B) is the correct option.