Step 1: Set up the counting problem differently.
Instead of listing triples by hand, treat each of the three directions as a bin that can hold any number of energy quanta, and ask how many quanta $n$ the second excited level carries.
Step 2: Find $n$ for the second excited state.
Level 0 (ground state) carries $n=0$ quanta, level 1 carries $n=1$, so the second excited level carries $n=2$ quanta above the ground state.
The energy of this level is $E = (n+\frac{3}{2})\hbar\omega$, so for $n=2$:
\[ E = \left(2 + \frac{3}{2}\right)\hbar\omega = \frac{7}{2}\hbar\omega \]
Step 3: Count the degeneracy with stars and bars.
The degeneracy is the number of ways to split $n=2$ identical quanta among 3 bins ($n_x, n_y, n_z$). This is a standard stars-and-bars count: choose 2 dividers among $n+2$ slots, giving $\binom{n+2}{2}$ arrangements.
\[ d = \binom{n+2}{2} = \binom{4}{2} = \frac{4!}{2!\,2!} = 6 \]
This counts the same six states as before ($(2,0,0)$ and permutations, $(1,1,0)$ and permutations) without listing them one by one.
Step 4: Rule out the distractors.
A degeneracy of 3 (option B) is what stars-and-bars gives for $n=1$, not $n=2$, since $\binom{3}{2}=3$. An energy of $\frac{5}{2}\hbar\omega$ (options C, D) belongs to $n=1$, one level too low.
Final Answer:
Matching energy $\frac{7}{2}\hbar\omega$ with degeneracy 6 fixes the answer as option (A).
\[ \boxed{E = \frac{7}{2}\hbar\omega,\ d = 6} \]