The concentration of donor atom is given by Nd

$N_d=N_ce^{-\left[\frac{E_C-E_F}{KT}\right]}$

Suppose that initially we have,

$N_d=x$ and $E_F=(E_F)_1$

Then

$x=N_ce^{-\left[\frac{E_C-(E_F)_1}{KT}\right]}$.....(1)

After doubling the concentration of donor atoms-

$N_d=2x$ and $E_F=(E_F)_2$

$2x=N_ce^{-\left[\frac{E_C-(E_F)_2}{KT}\right]}$.....(2)

Divide 1 and 2;

$\dfrac{x}{2x}=\dfrac{N_ce^{-\left[\frac{E_C-(E_F)_1}{KT}\right]}}{N_ce^{-\left[\frac{E_C-(E_F)_2}{KT}\right]}}$

$2=e^{\frac{-E_C+(E_F)_2+E_C-(E_F)_1}{KT}}$

$2=e^{\frac{0.4-E_C+(E_F)_2}{KT}}$

∵ $E_F-(E_F)_1=0.4$eV(data)

Taking log on both sides;

$log2=\dfrac{0.4-E_C+(E_F)_2}{KT}$

$\dfrac{E_C-(E_F)_2}{KT}=0.4-0.3010=0.099$eV

$\therefore\ E_C-(E_F)_2=0.099$KT eV …