Answer
$0.121$
Work Step by Step
From the equation $E_F = [\frac{3}{16 \sqrt2 \pi}]^{2/3}\frac{(hc^3)}{m_e c^{2}} n^{2/3}$
The numerical factor of $ [\frac{3}{16 \sqrt2 \pi}]^{2/3}$ is approximately $0.121$
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