The relation to use. For a non-degenerate n-type semiconductor the electron concentration is tied to the position of the Fermi level below the conduction band edge by

\(n=N_c\exp\left(-\dfrac{E_c-E_F}{kT}\right)\)

Inverting it gives exactly what the question asks for — the distance of E_F from the bottom of the conduction band:

\(E_c-E_F=kT\ln\left(\dfrac{N_c}{n}\right)\)

Step 1 — identify n. Arsenic is a group-V element, so it is a donor in silicon. At room temperature the donors are fully ionised, so every donor gives one conduction electron:

\(n\approx N_D=10^{16}\ \text{cm}^{-3}\)

Step 2 — form the ratio.

\(\dfrac{N_c}{n}=\dfrac{2.8\times10^{19}}{1\times10^{16}}=2800\)

Step 3 — take the logarithm and multiply by the thermal voltage. At 300 K, \(kT=0.0259\ \text{eV}\):

\(\ln(2800)=7.937\)

\(E_c-E_F=0.0259\times7.937=0.2056\ \text{eV}\)

which rounds to 0.206 eV, option 2.

A sanity check on the position. Silicon's gap is 1.12 eV, so mid-gap sits 0.56 eV below E_c. Our answer places E_F only 0.21 eV below the conduction band — well above mid-gap, which is exactly where the Fermi level must sit in n-type material. Had the answer come out beyond 0.56 eV the doping type would have been wrong.

How the doping level moves E_F. The dependence is logarithmic, so it takes a large change in doping to move the Fermi level appreciably:

Every factor of ten in doping shifts E_F by only \(kT\ln 10=0.06\) eV. Push the doping high enough and E_F enters the conduction band itself, at which point the material is degenerate and this Boltzmann formula no longer applies.

Hence, the Fermi level lies 0.206 eV below the bottom of the conduction band.