What is the electron configuration using core notation for ${\mathrm{Se}}^{2-}$?

Selenium belongs to group 16 and period 4 of the periodic table. It is represented by Se and the atomic number of selenium is 34. 

Selenium belongs to p block of the periodic table as the last electron is present in its p subshell. As the atomic number is 34 so number of protons in selenium is 34 and for an element to be electrically neutral the number of protons and electrons in that element are equal so number of electrons in selenium in the ground state is 34.

Since selenium has charge 2- that is electrons are in excess that is selenium has gain 2 electrons so the number of electrons in selenium are 34 + 2 = 36.

Electronic configuration of Se in Normal Notation:$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^4$. If the element lose or gain an electron it will do it with its outermost electrons that is with p electrons, so selenium gains 2 electrons and accommodates it in p subshell.

So, the electronic configuration of Se²⁻ in Normal Notation: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^6$
Hence, option (A) is correct.