The 8085 has a 16 bit address bus.
The address bus in the 8085 is 16 bits wide.
8 bits
You cannot address 1GB memory with the 8085 or the 8086/8088 without some kind of external demultiplexor that is software controlled. The address bus on the 8085 is 16 bits, giving addressibility of 64KB; while the address bus on the 8086/8088 is 20 bits, giving addressibility of 1MB. To address 1GB, you need a 30 bit address bus.
The stack pointer is 16 bits in size on the 8085 because that is how Intel designed it. The address bus is also 16 bits, so it made sense for the program and stack to be located anywhere in that address space.
The memory capacity of the 8085 microprocessor is 64 kb because the address bus is 16 bits, and you can address 216, or 64kb, with a 16 bit address bus.
The 8085 is an 8-bit microprocessor. Even though there are some 16-bit registers (BC, DE, HL, SP, PC), with some 16-bit operations that can be performed on them, and a 16-bit address bus, the accumulator (A), the arithmetic logic unit (ALU), and the data bus are 8-bits in size, making the 8085 an 8-bit computer.
The 8085 is an 8 bit processor, so its word length is 8 bits.
64 kb
The width of the address bus on the 8085 was decided based on a compromise between functionality, cost, and complexity. Intel decide to use 16 bits, as that was in keeping with common design at that time, and also because the 8085 was actually an enhanced version of the 8080.
The data size in the 8085 is 8 bits.
It depends on the size of the address bus, which is often different than the size of the data bus.If the address bus were 8 bits, then you could address 256 locations.If the address bus were 16 bits, such as in the 8085, then you could address 65,536 locations.If the address bus were 20 bits, such as in the 8086/8088, then you can address 1,048,576 locations.
8085 is a 8 bit microprocessor and so A register which is also known as accumulator is also 8 bit.