= Solution
The <Born rule> gives the required <probability> by integrating the <probability density> over the central interval. Put $d=0.1\,\mathrm{nm}$. For this <infinite square well> <ground state>,
$$
P(|x|\leq d)=\frac2L\int_{-d}^{d}\cos^2\left(\frac{\pi x}L\right)dx
=\frac1L\int_{-d}^{d}\left[1+\cos\left(\frac{2\pi x}L\right)\right]dx
=\frac{2d}L+\frac1\pi\sin\left(\frac{2\pi d}L\right).
$$
Substitution of $d/L=0.1$ gives
$$
\boxed{P=0.2+\frac{\sin(0.2\pi)}\pi\simeq0.3871\simeq38.7\%.}
$$
This exceeds the interval's $20\%$ fraction of the box length because the <ground state> <probability density> is largest at the centre; it is not a uniform position distribution.
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