Answer
The element of the excited-state configuration is niobium ($Nb$). Its ground-state condensed electron configuration is $$Nb: [Kr]5s^24d^{3}$$
Work Step by Step
$[Kr]5s^24d^25p^1$
In the excited state, normally there is no loss or addition of electrons. Therefore, to find out which element it is, we still calculate the number of electrons and deduce the atomic number from there.
Looking at the noble gas $[Kr]$, we see there are 36 inner-shell electrons.
Overall, there are $36+2+2+1=41$ electrons. So, the atomic number of the element is 41. This element is niobium ($Nb$).
The nearest noble gas of the lower atomic number is $Kr$. Since, $Nb$ has 41 electrons, its ground-state condensed electron configuration is $$Nb: [Kr]5s^24d^{3}$$
