The Bohr model was extended by Arnold Sommerfeld. Sommerfeld was able to predict all the Hydrogen atom quantum numbers. This theory is referred to today by the name
"Old quantum theory". Please see also, Hazhar Ghaderi's Bachelor thesis.
The basic idea of this theory is that closed classical trajectories (not necessarily circular) lead to discrete quantum numbers. The quantization is obtained through modified condition, now known by the name "Bohr-Sommerfeld condition"
The application of this theory becomes difficult for many electron atoms where exact classical solutions are not available.
This theory was abandoned after the discovery of the Schroedinger equation which gave precise predictions for many electron atoms spectra.
Now, we know that the old quantum theory is only an approximation to the correct quantum theory which is valid in general for large quantum numbers (First order in the WKB quantization).
However, for integrable systems, the Bohr-Sommerfeld is known to be exact or almost exact.
However, this is not the end of the story. In 1980 Sniatycki discovered a deep geometrical meaning of the Bohr-Sommerfeld condition within the theory of geometric quantization.
Sniatycki's work is described briefly in Matthias Blau's lecture notes.
Heuristically Sniatycki's result states that there exists a coordinate system in which the wave functions are concentrated on subspaces of the phase space where the quantum numbers are fixed to their quantized values. These spaces are called "Bohr-Sommerfeld varieties".
Nowdays, This subject is under active research and was applied sussessfully to complicated quantization problems such as the quantization of the moduli space of flat connections.
Which is wrong, my example or the answer in the textbook?
Your example is wrong. You have two active electrons in the $p$ shell, and their total spin must couple either to $S=0$ or $S=1$, which correspond to singlet ($2S+1=1$) or triplet ($2S+1=3$) states. The target state you've been given is a doublet state (indicated by the $2S+1=2$ superscript), so you've already missed the mark.
More generally, if you want a doublet state (with $S=1/2$), then you need an odd number of electrons, since even numbers of electrons always have integer-valued total spin.
This then puts you into trouble, because having $n=2$ limits you to having only $p$ electrons with $\ell=1$ contributing to the orbital angular momentum, and if you have an odd number of such electrons, then you're restricted to an odd-integer value for $L$. This then completely eliminates the possibility of any $2 \ {}^2\mathrm{D}_J$ state, whatever the $J$.
(If any of the above is unfamiliar, then it's almost certainly because of an incomplete preparation in the quantum-mechanical procedure for adding angular momenta. This is a large and complex topic, and you should take it from the ground up.)
As for your more general question,
I want to confirm that whether the total orbital angular momentum quantum number $L$ is always less than $n$.
No, this is not the case (at least, for excited states). With a half-filled shell, say, on atomic nitrogen, it's perfectly possible to achieve $\rm F$ states with $L=3$, by taking the parallel configuration for the three individual orbital angular momenta.
Best Answer
The principle quantum number is denoted $n$ because it is a 'natural number', $n=1,2,3,4....$.
The secondary azimuthal quantum number (also known as orbital angular momentum) is denoted $\ell$ through its association with angular momentum (typically denoted $L$).