The coefficient of static friction between a wooden block of mass 0 . 5   kg and a vertical rough wall…

$\mathrm{F}.\mathrm{B}.\mathrm{D}.$ of the block is shown in the diagram

Since block is at rest therefore

$fr-mg=0 ...\left(1\right)$

$F-N=0 ...\left(2\right)$

$fr\le \mu N$

In limiting case

$fr=\mu N=\mu F \dots \left(3\right)$

Using eq. $\left(1\right)$ and $\left(3\right)$

$\therefore \mu F=mg$

$\Rightarrow F=\frac{0.5\times 10}{0.2}=25 \mathrm{N}$