$F = qvB$ (Force of charged particle in a magnetic field)

And we know that

$F =\frac{ mv ^2}{ r } \quad$ (r is the radius of motion and $m$ is mass of particle)

$\Rightarrow qvB =\frac{ mv ^2}{ r }$

$\Rightarrow r =\frac{ mv }{ Bq }$

Now as we know that

$\omega=\frac{ V }{ r }$

$\Rightarrow \omega=\frac{ Bq }{ m }$

Time period, $T =\frac{2 \pi}{\omega}$

$\Rightarrow T =\frac{2 \pi m }{ Bq }$

And this shows that it is independent of both radius and velocity.