1. Given data

• Magnification: $m=-\dfrac{1}{3}$
• Distance between object and image: $30\text{ cm}$
• Need to find focal length $f=\dfrac{x}{4}$ cm, then report $|x|$

2. Use mirror magnification formula

For mirrors, $m=-\frac{v}{u}$

Given $-\frac{v}{u}=-\frac{1}{3}$ So, $\frac{v}{u}=\frac{1}{3}$ $v=\frac{u}{3}$

3. Use distance between object and image

Since magnification is negative, image is real and inverted. For a mirror, real object and real image lie on the same side of the mirror, so the separation is $|u-v|=30$

Now substitute $v=\dfrac{u}{3}$: $\left|u-\frac{u}{3}\right|=30$ $\frac{2|u|}{3}=30$ $|u|=45$

Using Cartesian sign convention for a real object in front of mirror, $u=-45\text{ cm}$ Then $v=\frac{u}{3}=-15\text{ cm}$

4. Apply mirror formula

Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Substitute $u=-45$ and $v=-15$: $\frac{1}{f}=\frac{1}{-15}+\frac{1}{-45}$ $\frac{1}{f}=-\frac{1}{15}-\frac{1}{45}$ $\frac{1}{f}=-\frac{3+1}{45}=-\frac{4}{45}$

Hence, $f=-\frac{45}{4}\text{ cm}$

5. Relate with given form

Given, $f=\frac{x}{4}\text{ cm}$ So, $\frac{x}{4}=-\frac{45}{4}$ $x=-45$ Therefore, $|x|=45$

6. Final answer

The required integer is $\boxed{45}$