Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
To normalize the wave function $ \psi(x) $, the total probability of finding the electron must be 1. This means:
$$ \int_{-\infty}^{\infty} |\psi(x)|^2 \; dx = 1 $$
Since $ \psi(x) = 0 $ for $ x < 0 $ or $ x > L $, we only need to integrate from $ 0 $ to $ L $. The normalization condition is:
$$ \int_{0}^{L} \left|c \sin\left( \frac{2 \pi x}{L} \right)\right|^2 \; dx = 1 $$
$$ c^2 \int_{0}^{L} \sin^2\left( \frac{2 \pi x}{L} \right) \; dx = 1 $$
Using the integral identity for $ \sin^2(\theta) $, where $ \int \sin^2(\theta) \; d\theta = \frac{\theta}{2} - \frac{\sin(2\theta)}{4}
$
Thus,
$$ c^2 \left[ \frac{x}{2} - \frac{\sin\left( \frac{4 \pi x}{L} \right)}{4 \pi/L} \right]_{0}^{L} $$
Hence,
$$ c^2 \cdot \frac{L}{2} = 1 $$
Thus:
$$ c^2 = \frac{2}{L} $$
$$ \boxed{c =\color{red}{\sqrt{\frac{2}{L}}}} $$
$$\color{blue}{\bf [b]}$$
The wave function $ \psi(x) $ is given by:
$$
\psi(x) = \begin{cases}
\sqrt{\frac{2}{L}} \sin\left( \frac{2 \pi x}{L} \right) & \text{for} \; 0 \leq x \leq L \\\\
0 & x\lt0,\;{\rm or}, x\gt L
\end{cases}
$$
$$\color{blue}{\bf [c]}$$
The probability density $ |\psi(x)|^2 $ is the square of the wave function:
$$
|\psi(x)|^2 = \left[\sqrt{\frac{2}{L}} \sin\left( \frac{2 \pi x}{L} \right) \right]^2 = \frac{2}{L} \sin^2\left( \frac{2 \pi x}{L} \right)
$$
$$\color{blue}{\bf [d]}$$
The probability is given by the integral of $ |\psi(x)|^2 $ over the interval $ 0 \leq x \leq L/3 $:
$$ P\left( 0 \leq x \leq \frac{L}{3} \right) = \int_{0}^{L/3} |\psi(x)|^2 \; dx $$
Substituting $ |\psi(x)|^2 = \frac{2}{L} \sin^2\left( \frac{2 \pi x}{L} \right) $,
$$ P \left( 0 \leq x \leq \frac{L}{3} \right)= \frac{2}{L} \int_{0}^{L/3} \sin^2\left( \frac{2 \pi x}{L} \right) \; dx $$
Using the identity of $ \sin^2(\theta) = \dfrac{1 - \cos(2\theta)}{2}$
$$ P \left( 0 \leq x \leq \frac{L}{3} \right)= \frac{1}{L} \int_{0}^{L/3} \left[ 1 - \cos\left( \frac{4 \pi x}{L} \right) \right] \; dx $$
$$ P \left( 0 \leq x \leq \frac{L}{3} \right)= \frac{1}{L} \left[ x - \frac{L}{4\pi} \sin\left( \frac{4\pi x}{L} \right) \right]_{0}^{L/3} $$
$$ P \left( 0 \leq x \leq \frac{L}{3} \right)= \frac{1}{L} \left[ \frac{L}{3} - \frac{L}{4\pi} \sin\left( \frac{4\pi}{3} \right) \right] $$
$$ P \left( 0 \leq x \leq \frac{L}{3} \right)= \frac{1}{3} - \frac{1}{4\pi} \sin\left( \frac{4\pi}{3} \right) $$
$$ P\left( 0 \leq x \leq \frac{L}{3} \right) = 0.4022=\color{red}{\bf 40.2\%} $$
