As shown in the figure a block of mass 10   kg lying on a horizontal surface is pulled by a force F…

The free body diagram for the given scenario is shown below-

From the equilibrium of the vertical component of forces, it can be written that

$\begin{array}{rcl}N& =& mg-F \sin 30°\\ & =& mg-\frac{F}{2}\\ & =& 100-\frac{F}{2}\\ & =& \frac{200-F}{2}........................\left(1\right)\end{array}$

From the equilibrium of horizontal component of forces, it can be written that,

$F \cos 30°= \mu N......................\left(2\right)$

Substitute the expression for the normal reaction force from equation (1) into equation (2) and solve to calculate the value of the applied force.

$\begin{array}{rcl}\sqrt{3}\frac{F}{2}& =& 0.25\times \left(\frac{200-F}{2}\right)\\ 4\sqrt{3}F& =& 200-F\\ F& =& \frac{200}{4\sqrt{3}+1}\\ & \approx & 25.2\end{array}$