Answer
$0.462\;\rm cm$
Work Step by Step
First of all, we need to find the focal lengths of the two lenses, where the magnification of the objective lens is given by
$$m_{\rm obj}=\dfrac{-L}{f_{\rm obj}}$$
So,
$$f_{\rm obj}=\dfrac{-L}{m_{\rm obj}}=\dfrac{18.0}{40}=\bf 0.45\;\rm cm$$
we neglected the negative sign since we are dealing now with lengths.
The magnification of the eye lens is given by
$$M_{\rm eye}=\dfrac{25\;\rm cm}{f_{\rm eye}} $$
So,
$$f_{\rm eye}=\dfrac{25\;\rm cm}{f_{\rm eye}}=\dfrac{25}{20}=\bf 1.25\;\rm cm$$
Now we have the focal length of the objective lens, and we can get the image position.
For a relaxed eye, the image of the objective lens must be at the focal point of the eye lens.
Thus,
$$s'=L-f_{\rm eye}=18-1.25=\bf 16.75\;\rm cm$$
Now we can easily find the position of the sample (the object) by using the thin lens formula.
$$\dfrac{1}{f_{\rm obj}}=\dfrac{1}{s}+\dfrac{1}{s'}$$
Hence,
$$s=\left[ \dfrac{1}{f_{\rm obj}}-\dfrac{1}{s'} \right]^{-1}$$
$$s=\left[ \dfrac{1}{0.45}-\dfrac{1}{16.75} \right]^{-1}$$
$$s=\color{red}{\bf 0.462}\;\rm cm$$
We can see that $s\approx f_{\rm obj}$
