Step 1: Understand the jump.
An electron in a hydrogen atom falls from the third excited state down to the ground state. We need the new de-Broglie wavelength compared with the old one.
Step 2: Link wavelength to the orbit number.
The de-Broglie wavelength is $\lambda = \frac{h}{p}$. In a Bohr orbit the electron's speed is proportional to $\frac{1}{n}$, so its momentum $p$ is also proportional to $\frac{1}{n}$. That makes $\lambda$ proportional to $n$.
Step 3: Find the two orbit numbers.
Ground state is $n = 1$. Counting up, first excited is $2$, second excited is $3$, and the third excited state is $n = 4$.
Step 4: Set up the ratio.
Since $\lambda \propto n$, we have $\dfrac{\lambda_{final}}{\lambda_{initial}} = \dfrac{n_{final}}{n_{initial}} = \dfrac{1}{4}$.
Step 5: Read what that means.
The electron goes from $n = 4$ to $n = 1$, so its wavelength shrinks to one quarter of the starting value.
Step 6: State the answer.
The new wavelength is $\frac{1}{4}$ of the original, which is option (B).
\[ \boxed{\lambda_{new} = \frac{1}{4}\,\lambda_{old}} \]