To determine the address space of a ROM IC interfaced to an 8085 microprocessor, we need to consider the size of the ROM and the addressing capability of the microprocessor.

Given that the ROM IC is 16K × 8, it means that the ROM has a storage capacity of 16 Kilobytes (K) and each memory location is 8 bits wide.

The 8085 microprocessor is an 8-bit microprocessor, which means it can address a maximum of 2^8 = 256 memory locations. However, it supports a 16-bit address bus, allowing it to access a maximum of 2^16 = 65,536 memory locations.

To calculate the address space, we need to determine the number of memory locations in the ROM IC. Since the ROM is 16K × 8, we can calculate the number of memory locations as follows:

16K = 16 * 1024 = 16,384 bytes
Each memory location is 8 bits, so the number of memory locations is 16,384 / 8 = 2,048

Therefore, the ROM IC has 2,048 memory locations.

Now we need to determine the address range for these memory locations. Since the microprocessor can access a maximum of 65,536 memory locations, the address range for the ROM IC will be from 0000H to FFFFH (in hexadecimal).

To convert this range into decimal, we have:
0000H = 0
FFFFH = 65,535

So, the ROM IC has memory locations from 0 to 65,535.

However, the ROM IC is a 16K × 8 ROM, which means its address space is limited to 16,384 memory locations. Therefore, the actual address space for the ROM IC is from 0000H to 3FFFH (in hexadecimal), which corresponds to memory locations 0 to 16,383 in decimal.

Hence, the correct answer is option 'B' - 0000H to 3FFFH.