= Solution
The closed $1s^2$ and $2s^2$ subshells contribute zero total <orbital angular momentum> and <spin>, so only the two $2p$ electrons matter. The first of <Hund's rules> maximizes their total spin, giving $S=1$. The <Pauli exclusion principle> then requires an antisymmetric spatial <wavefunction>. Two parallel-spin electrons cannot occupy the same $m_l$ orbital: the greatest possible orbital projection is $1+0=1$, so the maximal allowed $L$ within this triplet is $L=1$.
The ground term is therefore $^3P$, in <term symbol> notation $^{2S+1}L_J$. Since the $p^2$ subshell contains fewer than half of its six possible electrons, the third of <Hund's rules> places the smallest $J$ lowest. <Addition of angular momentum> gives $J=0,1,2$, in that energy order within the ground term. Thus
$$
\boxed{\text{ground level: }1s^2\,2s^2\,2p^2\;{}^3P_0\quad(L=1,S=1,J=0),}
$$
$$
\boxed{\text{first two excited levels: }{}^3P_1\ (J=1),\qquad{}^3P_2\ (J=2).}
$$
These are the <fine structure> partners of the same electronic term, with magnetic multiplicities $1,3,5$, rather than different principal-shell excitations. The ordering agrees with the tabulated lowest neutral-<carbon> levels.https://physics.nist.gov/PhysRefData/Handbook/Tables/carbontable5.htm[NIST, Energy Levels of Neutral Carbon, lowest three entries]{ref}
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