The ionic strength $\left(\mu \right)$of the $\text{0.1\hspace{0.17em}M}$or $\text{0.1\hspace{0.17em}m}$solution of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$is  $K$, then the value of $10K=$….

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{3Na}}^{+}+{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PO}}_4^{-3}$

Initial conc.    0.1m         0                        0     

Final conc.        0       $\text{3×0.1=0.3m}$       0.1m

Valences of ${\text{Na}}^{\text{+}}\text{=1,}$    molraity of ${\text{Na}}^{\text{+}}$= 0.3

Valences of ${\text{PO}}_{\text{4}}^{\text{-3}}\text{=3,}$     molraity of ${\text{PO}}_4^{-3}\text{ = 0.1}$ 

$\mu \left({\text{Na}}_{\text{3}}{\text{Po}}_{\text{4}}\right)\text{=}\frac{\text{1}}{\text{2}}\left({\text{c}}_{\text{1}}{\text{z}}_{\text{1}}^{\text{2}}{\text{+C}}_{\text{2}}{\text{z}}_{\text{2}}^{\text{2}}\right)$

                      $=\frac{1}{2}\left(0.3{\left(1\right)}^2+0.1{\left(3\right)}^2\right)$

                     =0.6