= Solution
In the isotropic parabolic-band approximation implied by the single supplied curvature, the <hole mass from a valence-band maximum> is
$$
m_h^*=\frac{\hbar^2}{|E''(0)|}=4.56\times10^{-31}\,\mathrm{kg}\simeq0.500\,m_e.
$$
The <hole> has charge $+e$, so its cyclotron angular frequency is
$$
\boxed{\omega_c=\frac{eB}{m_h^*}=3.52\times10^{11}\,\mathrm{rad\,s^{-1}}.}
$$
If “frequency” means cycles per second, \b[$f_c=\omega_c/(2\pi)=5.59\times10^{10}\,\mathrm{Hz}$]. The curvature unit is $\mathrm{J\,m^2}$, so $\hbar^2/E''$ has mass units. The external constants used are $\hbar=1.054571800\times10^{-34}\,\mathrm{J\,s}$, $e=1.6021766208\times10^{-19}\,\mathrm C$ and $m_e=9.10938356\times10^{-31}\,\mathrm{kg}$.<CODATA recommended values of the fundamental physical constants 2014 by National Institute of Standards and Technology>[Reduced Planck constant, elementary charge and electron mass]{ref} A genuinely anisotropic band would require the cyclotron mass for the orbit plane, not a single arbitrary directional curvature.
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