Answer
$0.10$
Work Step by Step
The normalized wavefunction $\psi_n(x)$ for an electron in an infinite, one-dimensional potential well with length $L$ along an $x$ axis are given by
$\psi_n(x)=\sqrt {\frac{2}{L}}\sin\Big(\frac{n\pi}{L}x\Big),\;\;\;\text{for}\;n=1,2,3....,$
where $n$ is the quantum number.
Now, the probability that the electron will be detected in the interval between $x$ and $x+dx$ inside of this well is given by
$p(x)=\psi^2_n(x)dx$
or, $p(x)={\frac{2}{L}}\sin^2\Big(\frac{n\pi}{L}x\Big)dx$
In our case, the electron trapped in the potential well having $L=100\;pm$ is in its ground state $(n=1)$. The electron can be detected in an interval of width $\Delta x=5.0\;pm$. The interval $\Delta x$ is so narrow that we can take the probability density to be constant within it.
Therefore, the probability of finding electron in an interval $\Delta x=5.0\;pm$ centered at $x=50\;pm$ is
$p(50)={\frac{2}{100\;pm}}\sin^2\Big(\frac{\pi}{100}50\Big)5\;pm=0.10$
