Given that, number of observations, $n=5$
$\begin{aligned}
& a_1=2.4 \mathrm{~m}, a_2=2.5 \mathrm{~m}, a_3=2.6 \mathrm{~m}, a_4=2.8 \mathrm{~m}, \\
& a_5=3.0 \mathrm{~m}
\end{aligned}$
Mean value of observations,
$\begin{aligned}
\bar{a} & =\frac{a_1+a_2+a_3+a_4+a_5}{5} \\
\bar{a} & =\frac{2.4+2.5+2.6+2.8+3.0}{5} \\
& =\frac{13.3}{5}=2.66
\end{aligned}$
Absolute errors in individual observed value
$\begin{aligned}
& \left|\Delta a_1\right|=\left|a_1-\bar{a}\right|=|2.4-2.66|=|-0.26|=0.26 \\
& \left|\Delta a_2\right|=\left|a_2-\bar{a}\right|=|2.5-2.66|=|-0.16|=0.16 \\
& \left|\Delta a_3\right|=\left|a_3-\bar{a}\right|=|2.6-2.66|=|-0.06|=0.06 \\
& \left|\Delta a_4\right|=\left|a_4-\bar{a}\right|=|2.8-2.66|=0.14 \\
& \left|\Delta a_5\right|=\left|a_5-\bar{a}\right|=|3.0-2.66|=0.34
\end{aligned}$
Mean absolute error,
$\begin{aligned}
\Delta \bar{a} & =\frac{\left|\Delta a_1\right|+\left|\Delta a_2\right|+\left|\Delta a_3\right|+\left|\Delta a_4\right|+\left|\Delta a_5\right|}{n} \\
\Delta \bar{a} & =\frac{0.26+0.16+0.06+0.14+0.34}{5}=\frac{0.96}{5} \\
\Delta \bar{a} & =0.192
\end{aligned}$
$\begin{aligned}
\text {Relative error } & =\frac{\text { Mean absolute error }}{\text { Mean value }} \\
& =\frac{\Delta \bar{a}}{\bar{a}}=\frac{0.192}{2.66}=0.072
\end{aligned}$