1. Identify the forces on the block

A wooden block of mass $m = 0.5\,\text{kg}$ is pressed against a vertical rough wall by a horizontal force $F$.

Forces acting on the block:

• Weight downward: $mg$
• Normal reaction from wall: $N$
• Static friction along the wall: $f_s$
• Applied horizontal force: $F$

Since the block is to adhere to the wall (not slide down), static friction must balance its weight.

2. Horizontal direction

The applied force presses the block into the wall, so in horizontal equilibrium:

$N = F$

3. Vertical direction

To prevent the block from sliding down,

$f_s = mg$

Maximum static friction is:

$f_{s,\max} = \mu_s N$

Given:

$\mu_s = 0.2, \quad m = 0.5\,\text{kg}, \quad g = 10\,\text{m/s}^2$

So,

$mg = 0.5 \times 10 = 5\,\text{N}$

For the block to just remain at rest,

$mg \le \mu_s N$

$5 \le 0.2\,N$

Since $N = F$,

$5 \le 0.2F$

$F \ge \frac{5}{0.2} = 25\,\text{N}$

4. Required force

The minimum horizontal force required is

$\boxed{25\,\text{N}}$

So the integer answer is:

$\boxed{25}$