The company wants to check if the mean has changed. So, null and alternative hypothesis are given by
$$
H_0: \mu=2.4, \quad \mu \neq 2.4
$$
Define a test statistic $T$ as $T=\bar{X}$.
Test: reject the null hypothesis if $|\bar{X}-2.4|>c$.
By CLT, we can say that $\frac{\bar{X}-2.4}{0.1 / \sqrt{100}}=\frac{\bar{X}-2.4}{1 / 100} \sim \operatorname{Normal}(0,1)$.
Now,
$$
\begin{aligned}
\alpha & =P(|\bar{X}-2.4|>c) \\
\Longrightarrow 0.05 & =P\left(\left|\frac{\bar{X}-2.4}{1 / 100}\right|>\frac{c}{1 / 100}\right) \\
\Longrightarrow 0.05 & =P(|Z|>100 c) \\
\Longrightarrow 0.05 / 2 & =P(Z<-100 c) \\
\Longrightarrow-1.96 & =-100 c \Longrightarrow c=0.0196
\end{aligned}
$$
Since $|\bar{X}-2.4|=|2.35-2.4|=0.05>c$, reject $H_0$.