Step 1: Recall the speed rule.
In the Bohr model, the speed of the electron in the $n$-th orbit is $v_n = \dfrac{Z e^2}{2 \varepsilon_0 h n}$. The only thing that changes from orbit to orbit here is the $n$ in the bottom, so speed falls as $n$ rises.
Step 2: Write it as a simple proportion.
Since everything else stays fixed for the same atom, we can say $v_n \propto \dfrac{1}{n}$, or simply $v_n = \dfrac{v_1}{n}$.
Step 3: Put in the first orbit value.
We are told the speed in the first orbit is $x$. So $v_1 = x$.
Step 4: Move to the fourth orbit.
Here $n = 4$, so \[ v_4 = \frac{v_1}{4} = \frac{x}{4}. \]
Step 5: Sense check.
A higher orbit means the electron sits farther out and moves slower. Going from $n=1$ to $n=4$ should cut the speed a lot, and dividing by 4 fits that picture.
Step 6: State the answer.
The speed in the fourth orbit is one quarter of the first.
\[ \boxed{x/4} \]