Answer
(a) Fluorine in the ground state.
(b) Gallium in the ground state.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have $ 1s^2 \, 2s^2 \, 2p^5 $ where
$ 1s^2 $: 2 electrons
$ 2s^2 $: 2 electrons
$ 2p^5 $: 5 electrons
So, the total Electrons: $ 2 + 2 + 5 =\bf 9 \;\rm e^-$
The element with 9 electrons is $\underline{\color{red}{\text{ Fluorine (F)}}}$, which has an atomic number of $Z=9$.
$\Rightarrow$ The configuration $ 1s^2 \, 2s^2 \, 2p^5 $ follows the expected electron filling order for fluorine, according to the Aufbau principle.
Therefore, this configuration represents $\underline{\color{red}{\text{ the ground state for fluorine}}}$.
$$\color{blue}{\bf [b]}$$
We have $ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^1 $ where
$ 1s^2 $: 2 electrons
$ 2s^2 $: 2 electrons
$ 2p^6 $: 6 electrons
$ 3s^2 $: 2 electrons
$ 3p^6 $: 6 electrons
$ 4s^2 $: 2 electrons
$ 3d^{10} $: 10 electrons
$ 4p^1 $: 1 electron
So, the total Electrons: $ 2 + 2 + 6 + 2 + 6 + 2 + 10 + 1 = \bf 31\;\rm e^- $
The element with 31 electrons is $\underline{\color{red}{\text{ gallium (Ga)}}}$, with an atomic number of $Z=31$.
The usual ground-state configuration for gallium is indeed $ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^1 $. Thus, this configuration represents$\underline{\color{red}{\text{ the ground state for gallium}}}$.
