Free body diagram (FBD) of the block (shown by a dot) is shown in figure.




For vertical equilibrium of the block,

$N=mg+F\sin 60°=\sqrt{3}g+\sqrt{3}\frac{F}{2} \dots .\left(\mathrm{i}\right)$

For no motion, force of friction

$f\ge F\cos 60°$

$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }°\mathrm{\mu } N\ge F\cos 60°$

$\mathrm{o}\mathrm{r}\frac{1}{2\sqrt{3}}\left(\sqrt{3}\mathrm{ }g+\frac{\sqrt{3} F}{2}\right)\ge \frac{F}{2}$

$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }g\ge \frac{F}{2} \mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }F\le 2 g \mathrm{o}\mathrm{r}\mathrm{ }20\mathrm{ }\mathrm{N}$

Therefore, maximum value of F is 20 N.