= Solution
Let $\mathcal A$ be the total two-dimensional area. There are $\mathcal A/(2\pi)^2$ <wavevector> states per reciprocal-space area per <spin>. Including <spin degeneracy> two and filling a disk gives
$$
N_e=2\frac{\mathcal A}{(2\pi)^2}\pi k_F^2
=\frac{\mathcal A k_F^2}{2\pi}.
$$
For areal <number density> $n_e=N_e/\mathcal A$ this yields
$$
\boxed{k_F=\sqrt{2\pi n_e},\qquad E_F=\frac{\pi\hbar^2 n_e}{m_e}.}
$$
The disk is centred at zero <wavevector>, and its boundary is a <fermi circle>. Omitting the two <spin> states would incorrectly double the <energy> inferred for a given <number density>.
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