= Solution
Treat the metal gate and conducting electron sheet as a parallel-plate capacitor separated by a uniform GaAs-like dielectric. Its capacitance per area is
$$
C_g/A=\frac{\epsilon_r\epsilon_0}{d}
=\frac{12.9(8.85\times10^{-12})}{200\times10^{-9}}
=5.71\times10^{-4}\,\mathrm{F\,m^{-2}}.
$$
A gate-voltage change gives an electron-number change of magnitude
$$
\boxed{|\Delta n_s|=\frac{(C_g/A)|\Delta V_g|}{e}
=3.56\times10^{15}\,\mathrm{m^{-2}}
=3.56\times10^{11}\,\mathrm{cm^{-2}}.}
$$
For the ordinary accumulation convention, a more positive gate voltage attracts more electrons, while a more negative gate depletes them. The magnitude is about $13\%$ of the preceding sheet density.
This neglects depletion, trapped charge, finite well width and <quantum capacitance>. The occupied-subband quantum capacitance is $C_q/A=e^2D_{2D}\simeq0.0449\,\mathrm{F\,m^{-2}}$, much larger than $C_g/A$ here; putting the two in series would reduce the estimate by only about $1.3\%$ within this simple model.
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