Between $4s$ and $3d$ orbitals,which one has higher energy and why?

(B) The energy of the $3d$ orbital is higher than that of the $4s$ orbital.
This is determined by the $(n+l)$ rule:
For $4s$: $(n+l) = (4+0) = 4$.
For $3d$: $(n+l) = (3+2) = 5$.
Since the $(n+l)$ value for $3d$ is greater than that for $4s$,the $3d$ orbital has higher energy.