Answer
$$ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}$$
or
$$ [Kr]4d^{10}$$
Explanation:
The atomic number of cadmium, Cd, is 48. Thus, the ground state electron configuration of this element is
$$ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}$$
Or
$$ [Kr] 4d^{10}5s^{2}$$
But since the problem is draw the electron configuration of the $$ Cd^{2+ } $$ his means that the element lost $$ 2e^{-}$$
Thus, the electron configuration of $$ Cd^{2+}$$ is
$$ 1s^{2} 2s^{2} 2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10} $$