Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the normalization condition requires that the total probability of finding the electron over the entire space is equal to 1. So,
$$ \int_{-\infty}^{\infty} |\psi(x)|^2 \; dx = 1 $$
Since the wave function is 0 for $ |x| \geq 1 \; \text{cm} $, the integral only needs to be evaluated from $ x = -1 \; \text{cm} $ to $ x = 1 \; \text{cm} $
$$ \int_{-1}^{1} |c \sqrt{1 - x^2}|^2 \; dx = 1 $$
$$ c^2 \int_{-1}^{1} (1 - x^2) \; dx = 1$$
The integral can be split into two parts;
$$c^2\left[ \int_{-1}^{1} 1 \; dx - \int_{-1}^{1} x^2 \; dx \right]=1$$
where $ \int_{-1}^{1} 1 \; dx = 2$, and $ \int_{-1}^{1} x^2 \; dx = \frac{2}{3}$
Thus,
$$ c^2 \left[2 - \frac{2}{3} \right] = 1 $$
$$ c^2 \cdot \frac{4}{3} = 1 $$
$$ c^2 = \frac{3}{4} $$
Therefore,
$$ c = \color{red}{\bf \frac{\sqrt{3}}{2}}\;\rm cm^{-1/2}$$
$$\color{blue}{\bf [b]}$$
The wave function $ \psi(x) $ is given by
$$ \psi(x) = \begin{cases}
\frac{\sqrt{3}}{2} \sqrt{1 - x^2} & \text{for} \; |x| \leq 1 \; \text{cm} \\
0 & \text{for} \; |x| > 1 \; \text{cm}
\end{cases} $$
$$\color{blue}{\bf [c]}$$
The probability density $ |\psi(x)|^2 $ is given by
$$|\psi(x)|^2 = \begin{cases}
\left( \frac{\sqrt{3}}{2} \sqrt{1 - x^2} \right)^2 & \text{for} \; |x| \leq 1 \; \text{cm} \\
0 & \text{for} \; |x| > 1 \; \text{cm}
\end{cases} $$
Simplifying:
$$ |\psi(x)|^2 = \frac{3}{4} (1 - x^2) \quad \text{for} \; |x| \leq 1 \; \text{cm} $$
$$\color{blue}{\bf [d]}$$
We know that
$$\text{Prob}( \text{in } \delta x \text{ at } 0\leq x\leq 0.5) =\text{Area under the $|\psi(x)|^2$ curve}\\
=\dfrac{N}{N_{tot}}$$
where $N$ is the number of detected electrons while $N_{tot}$ is the total number of electron.
Thus,
$$N=N_{tot}\cdot \text{Prob}( \text{in } \delta x \text{ at } 0\leq x\leq 0.5) $$
$$N=N_{tot} \int_{0.00 \, \text{cm}}^{0.50 \, \text{cm}} |\psi(x)|^2 dx$$
$$N= \frac{3}{4}N_{tot} \int_0^{0.50} (1 - x^2) dx$$
$$N = \frac{3}{4}(10^5) \left[ x - \frac{x^3}{3} \right]_0^{0.50} $$
$$N=\color{red}{\bf 3437}\;\rm electron$$
