What is the position of the element in the periodic  table satisfying the electronic configuration

$(n-1)d^{1}n{s}^2\text{ for }n=4?$

Here, the outermost electron enters the 3rd shell, which causes the element to belong to the d-block. Since the number of electrons in the (n-1) orbital + the number of electrons in the n orbital = 2+1=3, the element is a member of the 3rd group. Since the element's period, in this case, equals the principal quantum number. So, the element belongs to the fourth period.

Hence, the correct answer is D.