= Solution
Use the magnitudes of the single-electron-band Drude Hall relation and longitudinal sheet resistivity:
$$
|\rho_{xy}|=\frac{B}{n_se},\qquad \rho_{xx}=\frac1{n_se\mu}.
$$
The given values are two-dimensional resistivities, so no sample thickness is inserted. Thus
$$
\boxed{n_s=\frac1{(1.602\times10^{-19})(2083)}
=2.997\times10^{15}\ \mathrm{m^{-2}}=2.997\times10^{11}\ \mathrm{cm^{-2}},}
$$
$$
\boxed{\mu=\frac{|\rho_{xy}|}{B\rho_{xx}}=\frac{2083}{0.6}
=3.472\times10^3\ \mathrm{m^2\,V^{-1}s^{-1}}
=3.472\times10^7\ \mathrm{cm^2\,V^{-1}s^{-1}}.}
$$
The carrier sign follows from a specified Hall-voltage orientation; these calculations use positive magnitudes. A parallel conducting layer, several occupied subbands or a quantized Hall plateau would require a different interpretation of a single-field measurement.
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