Floating Point Arithmetic with 3 bits mantissa
The problem is underspecified, but I'll assume IEEE 754 default behavior.
In IEEE 754, at least in space-constrained floating-point formats, a bit is saved by not storing the leading bit of the mantissa, which is always 1. So the representation of the number 1 would have 000 in the three mantissa bits.
The next largest representable number is 1.001 in binary (001 in the mantissa bits). The next smallest representable number I'll leave for you to figure out (it's not 0.111).
In IEEE 754, addition works by calculating the exact sum of the addends over the reals (infinite precision), then rounding to a representable floating-point number. In the default mode (round to even), you round to the nearest representable value, or if you're exactly halfway between representable values, the tie is broken in favor of the one with 0 in the least significant mantissa bit.
If $a$ is the next smallest representable number below 1, and $b$ is 1.001, then sums in the interval $\left[\frac{a+1}2, \frac{1+b}2\right]$ will be rounded to 1. (It's closed on both ends because round-to-even breaks the tie in favor of 1.) From there it's easy to answer the question.