Step-by-step Solution:

In an 8-bit 2's complement system, the range of integers is from (-128) to (127). This is because the range of 8 bits in 2's complement is given by: Range = \(-2^{n-1} \) to \(2^{n-1} - 1\) Where (n = 8): \(-2^7 = -128\) \(\quad\) and \(2^7 - 1 = 127\) So, the smallest integer representable by an 8-bit number in 2's complement is (-128). \[\] Correct Option: (b)