The total no. of electrons present in 48 g of $M{g}^{2+}$ are

No. of $M{g}^{2+}$ ions = $\frac{\text{Mass}}{\text{Gram}\text{ionic}\text{weight}}\text{x}{\text{N}}_{\text{A}}$

= $\frac{\text{48}}{\text{24}}{\text{N}}_{\text{A}}\text{=}{\text{2N}}_{\text{A}}$

No. of electrons in each $M{g}^{2+}$ ion = 10

Total no. of electrons = 10 $\times$ 2$N_A$  = 20 $N_A$