Step 1: The de Broglie wavelength shortcut.
For an electron accelerated through a potential difference, a handy formula gives the wavelength directly in angstrom: \[ \lambda = \frac{12.27}{\sqrt{V}}\ \text{\AA}, \] where $V$ is the accelerating voltage in volts.
Step 2: Track the electron's energy.
It enters plate $A$ already carrying $200\,\text{eV}$, and the plates add another $100\,\text{eV}$ as it crosses to $B$. So the total energy on exit is \[ E = 200 + 100 = 300\ \text{eV}. \]
Step 3: Match energy to an equivalent voltage.
An energy of $300\,\text{eV}$ behaves like acceleration through $V = 300\,\text{V}$ in the formula.
Step 4: Compute the wavelength.
\[ \lambda = \frac{12.27}{\sqrt{300}} = \frac{12.27}{17.32} \approx 0.71\ \text{\AA}. \]
Step 5: Compare with the choices.
This matches $0.713\,\text{\AA}$ closely.
Step 6: Conclusion.
The de Broglie wavelength on leaving plate $B$ is about $0.713\,\text{\AA}$. \[ \boxed{0.713\ \text{\AA}} \]