Answer
$1.926\times10^{28}\;m^{-3} (eV)^{-1}$
The above value is consistent with the curve of Fig. 41-6.
Work Step by Step
The density of states at energy $E$ is is
$N(E)=\frac{8\sqrt 2\pi m^{3/2}}{h^3}E^{1/2}$
or, $N(E)=CE^{1/2}$
Where,
$C=\frac{8\sqrt 2\pi m^{3/2}}{h^3}$ is a constant
Putting the known values, we obtain,
$C=\frac{8\times\sqrt 2\times\pi \times ( 9.109\times10^{31})^{3/2}}{( 6.626\times10^{-34} )^3}\;kg^{3/2}J^3 s^3$
or, $C=1.062\times10^{56}\;kg^{3/2}J^3 s^3$
or, $C=6.81\times10^{27}\;m^{-3} (eV)^{-3/2}$
Here, $E=8\;eV$
Therefore,
$N(E)=6.81\times10^{27}\;m^{-3}\times 8^{1/2}\;m^{-3} (eV)^{-1}$
or, $N(E)=1.926\times10^{28}\;m^{-3} (eV)^{-1}$
Thus, the value of $N(E)$ is consistent with the curve of Fig. 41-6.
