Magnetic, the moment of $X^{n+}(Z=26)\text{ is }\sqrt{24}$ B.M. Hence, the number of unpaired electrons and the value of n respectively are :

Magnetic moment $= \sqrt{\mathrm{n}(\mathrm{n} + 2)} =\sqrt{ 24 }$B.M. 
∴No. of unpaired electron = 4

$X_{26}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$

To get 4 unpaired electrons, the outermost configuration will be $3{\mathrm{d}}^6$

∴ No. of electrons lost = 2 (from $4{\mathrm{s}}^2$ ). 

∴ n = 2.

The correct option is A.