Step-by-step Solution:

1. Identify the forces acting on the proton. When the proton is released between the two vertical plates, it is subjected to two forces:

   • Gravitational Force ($F_g$): This force acts vertically downwards due to the Earth's gravity. It is given by $F_g = m_p g$, where $m_p$ is the mass of the proton and $g$ is the acceleration due to gravity.
   • Electric Force ($F_e$): The potential difference $X$ between the vertical plates creates a horizontal electric field $E$. This field exerts a horizontal force on the positively charged proton. The force is given by $F_e = eE$, where $e$ is the elementary charge.

2. Relate the electric field to the potential difference. For two large parallel plates with a separation $d$ and potential difference $X$, the magnitude of the uniform electric field $E$ between them is given by: $E = \frac{X}{d}$ Therefore, the electric force on the proton is: $F_e = e \frac{X}{d}$

3. Analyze the initial motion of the proton. The proton is released from rest. Its initial acceleration, and therefore its initial direction of motion, will be in the direction of the net force acting on it. The net force is the vector sum of the gravitational and electric forces.

   • The vertical component of acceleration is $a_y = \frac{F_g}{m_p} = \frac{m_p g}{m_p} = g$.
   • The horizontal component of acceleration is $a_x = \frac{F_e}{m_p} = \frac{eX}{m_p d}$.

4. Use the given angle to relate the forces. The problem states that the proton moves at an angle of $45^\circ$ to the vertical just after release. This means the angle $\theta$ that the net acceleration vector makes with the vertical is $45^\circ$. We can relate the components of acceleration using trigonometry: $\tan \theta = \frac{\text{horizontal component of acceleration}}{\text{vertical component of acceleration}} = \frac{a_x}{a_y}$ Given $\theta = 45^\circ$, we know that $\tan 45^\circ = 1$. Therefore: $1 = \frac{a_x}{a_y} \implies a_x = a_y$

5. Solve for the potential difference X. Substituting the expressions for $a_x$ and $a_y$: $\frac{eX}{m_p d} = g$ Now, we can rearrange this equation to solve for $X$: $X = \frac{m_p g d}{e}$

6. Substitute the known values and calculate X. We use the following standard values:

   • Mass of a proton, $m_p \approx 1.67 \times 10^{-27} \text{ kg}$
   • Acceleration due to gravity, $g \approx 9.8 \text{ m/s}^2$
   • Charge of a proton, $e = 1.6 \times 10^{-19} \text{ C}$
   • Separation between plates, $d = 1 \text{ cm} = 0.01 \text{ m}$

   Substituting these values into the equation for $X$: $X = \frac{(1.67 \times 10^{-27} \text{ kg}) \times (9.8 \text{ m/s}^2) \times (0.01 \text{ m})}{1.6 \times 10^{-19} \text{ C}}$ $X = \frac{1.67 \times 9.8 \times 0.01}{1.6} \times 10^{-27 - (-19)} \text{ V}$ $X = \frac{0.16366}{1.6} \times 10^{-8} \text{ V}$ $X \approx 0.1023 \times 10^{-8} \text{ V}$ $X \approx 1.023 \times 10^{-9} \text{ V}$

7. Compare the result with the given options. The calculated value $X \approx 1.023 \times 10^{-9} \text{ V}$ is very close to $1 \times 10^{-9} \text{ V}$.

   Therefore, the correct option is C.