Step 1: What decides the energy of an electron.
In a multi-electron atom the energy of an orbital is fixed by the $(n+l)$ rule, also called the Aufbau rule. A bigger $(n+l)$ value means a higher energy orbital.
Step 2: The tie-breaker rule.
If two orbitals have the same $(n+l)$ value, then the one with the larger $n$ has higher energy. So we just compute $(n+l)$ for each choice.
Step 3: Work out each option.
Option 1: $n=3, l=0 \Rightarrow n+l = 3$ (this is $3s$).
Option 2: $n=3, l=1 \Rightarrow n+l = 4$ (this is $3p$).
Option 3: $n=3, l=2 \Rightarrow n+l = 5$ (this is $3d$).
Option 4: $n=4, l=0 \Rightarrow n+l = 4$ (this is $4s$).
Step 4: Pick the largest.
The values are $3, 4, 5, 4$. The biggest is $5$, which belongs to Option 3, the $3d$ electron.
Step 5: Conclusion.
So the electron with the highest energy is the one with $n=3, l=2, m=1, s=+1/2$. \[ \boxed{n=3,\ l=2\ (3d)} \]