#### Answer

We take $\delta=0.2391$ here.

#### Work Step by Step

$$a=2.7591 \hspace{1cm} b=3.2391\hspace{1cm}c=3$$
Here we need to find $\delta\gt0$ such that for all $x$, $0\lt |x-3|\lt\delta$, then $2.7591\lt x\lt 3.2391$.
The sketch is shown below.
- From $3$ to $2.7591$: $3-2.7591=0.2409$
- From $3$ to $3.2391$: $3.2391-3=0.2391$
So the nearer endpoint of $(2.7591,3.2391)$ to $c=3$ is $3.2391$, and the distance is $0.2391$.
So if we take $\delta=0.2391$ or any smaller positive value, then $0\lt |x-3|\lt0.2391$, and $x$ will only range in $(2.7609,3.2391)$, as shown by the red lines in the sketch, thereby $x$ is always placed between $2.7591$ and $3.2391$ as requested.
In other words, $$0\lt |x-3|\lt0.2391\hspace{2cm}2.7591\lt x\lt 3.2391$$