$x\in Z$ but $x\ne 0$ where $x\in Z$

$-25\le x\le 25$

Now, $x+6\le \frac{135}{x}$

$\Rightarrow \frac{x^2+6x-135}{x}\le 0$

$\Rightarrow \frac{\left(x+15\right)\left(x-9\right)}{x}\le 0$

The solution set is $x\in (-\infty ,-15]\cup (0,9]$

Given that $\left\{x\in \mathrm{\mathbb{Z}}|-25\le x\le 25,x\ne 0\right\}$

Common interval is $x\in \left[-25,-15\right]\cup (0,9]$

Total number of integers in $(0, 9]$and $\left[-25 ,-15\right]$ are

$9+11=20$

Total numbers in sample space, $-25\le x\le 25, x\ne 0$are $50$

Favourable outcomes$=20$

Probability $=\frac{\mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} }{\mathrm{total} \mathrm{outcomes}}$

Probability$=\frac{20}{50}=\frac{2}{5}$