= Solution
Use physical fields proportional to the <real part> of $e^{-i\omega t}$. The <Drude frequency response> is
$$
\sigma(\omega)=\frac{\sigma_0}{1-i\omega\tau},\qquad \boxed{|\sigma(\omega)|^2=\frac{\sigma_0^2}{1+\omega^2\tau^2}.}
$$
Thus the sketch is an even <Lorentzian line shape>, with maximum $\sigma_0^2$ at zero <frequency>, half that value at $|\omega|=\tau^{-1}$, and an $\omega^{-2}$ tail. The positive-frequency half is shown in the preceding figure. Squaring the <modulus> differs from plotting either $|\sigma|$ or the <real part> of $\sigma$.
For a single carrier species, the <plasma frequency> and <electrical conductivity> satisfy
$$
\omega_p^2=\frac{nq^2}{m\epsilon_0},\qquad \sigma_0=\frac{nq^2\tau}{m}=\epsilon_0\omega_p^2\tau.
$$
Therefore
$$
\boxed{\omega_{1/2}=\frac1\tau=\frac{\epsilon_0\omega_p^2}{\sigma_0}=\frac{(8.854\times10^{-12})(5.6\times10^{15})^2}{9.9\times10^6}\simeq2.80\times10^{13}\,\mathrm{s^{-1}}.}
$$
The corresponding <relaxation time> is $\tau\simeq3.57\times10^{-14}\,\mathrm s$. This is an <angular frequency>; the ordinary <frequency> is $\boxed{f_{1/2}=\omega_{1/2}/(2\pi)\simeq4.46\times10^{12}\,\mathrm{Hz}}$. The units of <electrical conductivity> used in the calculation are $\mathrm{\Omega^{-1}m^{-1}}$, consistently with the SI plasma-frequency relation.
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