${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2\rightleftharpoons 3{\mathrm{Mg}}^{2+}+2{\mathrm{PO}}_4^{3-};K_{\mathrm{sp}}={\left[{\mathrm{Mg}}^{2+}\right]}^3{\left[{\mathrm{PO}}_4^{3-}\right]}^2$

lf x is the solubility of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$,then

$\begin{array}{r}\left(3x{)}^3\right(2x{)}^2=1.08\times {10}^{-13 }{\mathrm{M}}^5;x={\left(\frac{1.08\times {10}^{-13}}{108} {\mathrm{M}}^5\right)}^{1/5}=0.001 \mathrm{M}\\ \left[{\mathrm{Mg}}^{2+}\right]=3x=3\times {10}^{-3 }\mathrm{M}\end{array}$