An electron is observed executing uniform circular motion with a...

The frequency of revolution of an electron in a magnetic field is given by $f = \dfrac{qB}{2\pi m}$.

The magnetic field inside a long solenoid is $B = \mu_0 n I$, where $n$ is the number of turns per unit length.

Substituting $B$ into the frequency formula, we get:

$f = \dfrac{q \mu_0 n I}{2\pi m}$

Rearranging for $n$:

$n = \dfrac{2\pi m f}{q \mu_0 I}$

Substituting the standard values $m = 9.1 \times 10^{-31} \text{ kg}$, $q = 1.6 \times 10^{-19} \text{ C}$, $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$, and the given values $f = 1.00 \times 10^8 \text{ Hz}$, $I = 2.00 \text{ A}$:

$n = \dfrac{2\pi \times 9.1 \times 10^{-31} \times 10^8}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times 2}$

$n = \dfrac{9.1 \times 10^{-23}}{6.4 \times 10^{-26}}$

$n = 1421.875 \approx 1420 \text{ turns m}^{-1}$