1. Identify the orbitals

Given quantum numbers:

• A: $n=3,\ l=0 \Rightarrow 3s$
• B: $n=4,\ l=0 \Rightarrow 4s$
• C: $n=3,\ l=1 \Rightarrow 3p$
• D: $n=3,\ l=2 \Rightarrow 3d$

The magnetic quantum number $m$ does not affect energy for these orbitals in the usual ordering.

2. Use the $(n+l)$ rule

For multi-electron atoms, orbital energy increases with increasing value of $(n+l)$. If two orbitals have the same $(n+l)$ value, then the orbital with higher $n$ has higher energy.

Now calculate:

• For $3s$: $n+l=3+0=3$
• For $4s$: $n+l=4+0=4$
• For $3p$: $n+l=3+1=4$
• For $3d$: $n+l=3+2=5$

3. Arrange by decreasing energy

Highest $(n+l)$ means highest energy:

• $3d$ has $5$ $\Rightarrow$ highest
• $4s$ and $3p$ both have $4$
  • Between them, higher $n$ means higher energy
  • So $4s > 3p$
• $3s$ has $3$ $\Rightarrow$ lowest

Therefore, $3d > 4s > 3p > 3s$

4. Map back to letters

• $3d \Rightarrow D$
• $4s \Rightarrow B$
• $3p \Rightarrow C$
• $3s \Rightarrow A$

So the decreasing order is $D > B > C > A$

5. Check options

This corresponds to Option D.