The correct answer is C. 16 Mbytes.
The 80286 microprocessor has a 24-bit address bus, which means that it can address up to 2^24 = 16,777,216 bytes of memory. This is equivalent to 16 Mbytes.
Option A is incorrect because the 80286 microprocessor has a 24-bit address bus, not a 16-bit address bus.
Option B is incorrect because the 80286 microprocessor has a 24-bit address bus, not a 32-bit address bus.
Option D is incorrect because the 80286 microprocessor has a 24-bit address bus, not a 64-bit address bus.