= Solution
Treat the top gate as a <parallel-plate capacitor>, neglecting fringing and assuming its voltage change couples to the 2DEG without a dominant trap or series-capacitance correction. Its <capacitance> per area and induced <electron> density are
$$
\frac CA=\frac{\epsilon_0\epsilon_r}{d}=1.15102\times10^{-3}\ \mathrm{F\,m^{-2}},\qquad
\Delta n_s=\frac{\epsilon_0\epsilon_r\Delta V}{ed}
=1.43698\times10^{15}\ \mathrm{m^{-2}}.
$$
In this <top-gate density tuning>, a positive gate change increases the <electron> density in this model, giving $n_s'=4.43371\times10^{15}\ \mathrm{m^{-2}}$. Since $B_j\propto k_F\propto\sqrt{n_s}$,
$$
\boxed{|B'_1|=0.2197\ \mathrm T,\qquad |B'_2|=0.4395\ \mathrm T,\qquad |B'_3|=0.6592\ \mathrm T.}
$$
The fields rise by a factor $\sqrt{n_s'/n_s}=1.2163$, not by the density ratio itself. Finite <quantum capacitance> is a small correction here: the one-subband value $e^2m^*/(\pi\hbar^2)$ is much larger than the stated geometric <capacitance> per area.
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