= Solution
A step elastically reflects a surface-state wave. Interference between $e^{ikx}$ and $re^{-ikx}$ produces a local density oscillation at $2k$, so the measured spatial period is $\lambda_{\rm osc}=\pi/k$, \b[not] $2\pi/k$. Thus $k_1=\pi/20=0.15708\ \mathring{\mathrm A}^{-1}$ and $k_2=\pi/15=0.20944\ \mathring{\mathrm A}^{-1}$. Write $E=E_0+A k^2$ with $A=\hbar^2/(2m^*)$. The two points give
$$
A=\frac{0.50}{(\pi/15)^2-(\pi/20)^2}=26.054\ \mathrm{eV}\,\mathring{\mathrm A}^2,\qquad
\boxed{m^*=0.1462m_e}.
$$
Here $\hbar^2/(2m_e)=3.80998\ \mathrm{eV}\,\mathring{\mathrm A}^2$. The fitted minimum is $E_0=1.00-A(\pi/20)^2=\boxed{0.35714\ \mathrm{eV}}$ above the <Fermi level>. This assumes the bias measures sample excitation energy, <parabolic dispersion> over the measured interval, and no wave-vector-dependent phase contribution that changes the fitted period.
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