1. Identify the orbitals from the quantum numbers

The energy of an electron in a multi-electron atom depends mainly on:

• principal quantum number $n$
• azimuthal quantum number $l$

It does not depend on $m$ or $s$ for this comparison.

The given sets correspond to:

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

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 one with higher $n$ has higher energy.

Now calculate:

• A: $n+l = 3+0 = 3$
• B: $n+l = 3+1 = 4$
• C: $n+l = 3+2 = 5$
• D: $n+l = 4+0 = 4$

So the order of increasing energy is: $3s < 3p < 4s < 3d$

because between $3p$ and $4s$, both have $n+l=4$, and the orbital with larger $n$ ($4s$) has higher energy.

3. Find the highest energy

Among all options, the largest $(n+l)$ is for C: $3d$ Thus, C represents the highest energy.

4. Check each option

• A: $3s$ → not highest
• B: $3p$ → not highest
• C: $3d$ → highest
• D: $4s$ → lower than $3d$

5. Final answer

The correct option is: $\boxed{\text{C}}$