Consider a 32-bit processor of the Harvard Architecture that has a 4 KB (kilobyte) instruction memory. The memory address has 32 bits and each address identifies a single byte in the memory (e.g. address Ox00FF00E5 points to a single byte stored at that memory location). a) If the starting address of the first byte that can be stored in the instruction memory is 0x08000000, what is the final 32 bit address (in hexadecimal) of the instruction memory? (Note: 1 KB = 1024 bytes = 210) b) If the start and final address of the data memory is 0x20000000 and Ox2000FFFF, respectively, what is the size of this memory space in KB? Activate Windows