The forces acting on the block are shown. Since the block is not

moving forward for the maximum force F applied, therefore

 $F\cos 60°=f=\mu N$     …(i) (Horizontal Direction)

and  $F\sin 60°+mg=N$    …(ii)

From (i) and (ii)

 $F\cos 60°=\mu \left[F\sin 60°+mg\right]$

 $\Rightarrow F=\frac{\mu mg}{\cos 60°-\mu \sin 60°}=\frac{\frac{1}{2\sqrt{3}}\times \sqrt{3}\times 10}{\frac{1}{2}-\frac{1}{2\sqrt{3}}\times \frac{\sqrt{3}}{2}}=\frac{5}{\frac{1}{4}}=20\mathrm{N}$