Answer:
9 bits
Explanation:
We all recognize 1 K bytes as =1024 bytes.
And the storage medium used for program instructions are the non-volatile flash memories. The MCUs are always 8 bits, and a piece of the instructions revenues two or one 16-bit words. Besides we know program memory magnitude through the device name, and as an instance, the ATmega64x line emanates with 64KB of flash, and the ATmega32x line emanates with 32 KB. And there is nobody like off-chip program memory, and all of the codes that are being implemented by the AVR essentially be on the on-chip flash. However, this restraint is not ever applied to the AT94 FPSLIC AVR/FPGA chips. Also, you must know that AVR is a family of microcontrollers established by ATEL in 1996. It is a Harvard architecture type single-chip 8-bit RISC microcontroller. And it is supposed to be the first microcontroller which applied flash memory for storing the program They are heavily used in embedded systems.
1k byte = 1024 bytes, However, if the program instruction takes 16 bits.
And hence it only wishes 512.
Here and now 512 = 10^9, and henceforth you necessitate 9 bits at minimum.
excluding you can somehow paginate that (such as through one-bit pagination, you will be necessitating only 8 bits and so on.)
Therefore the answer here = 9 bits.
