Answer
$[Number \space of \space Occupied \space States] = 6.7 \times 10^{19}$
Work Step by Step
There are 4 steps in solving this question.
First we need to find the probability that a state of E occupied at temperature T.
$ P(E) = \frac{1}{e^{E_1 - E_F/kT} + 1}$
Then, solve the denominator first to make the next steps easier, Find the Fermi Energy in Joule
$E_F = \frac{(0.121)h^2}{m_e} n^{2/3}$
Where
$h = 6.626\times 10^{-34} J.s$
$m_e=9.109\times 10^{-31} kg$
n from the question $n = 1.70 \times 10^{28} m^{-3}$
$E_F = \frac{(0.121)(6.626\times 10^{-34} J.s)^2}{(9.109\times 10^{-31} kg)} (1.70 \times 10^{28} m^{-3})^{2/3}$
$E_F = 3.855 \times 10^{-19} J$
Take back the denominator to solve, where
$E$ in the question is $4.00 \times 10^{-19}J$
and $k = 1.38 \times 10^{-23} J/K$
and $T = 200k$
$e^{(4.00 \times 10^{-19}J) - (3.855 \times 10^{-19} J)/(1.38 \times 10^{-23} J/K)(200k) + 1}$
$ = 192.258$
So $ P(E) = \frac{1}{192.258}$
So $ P(E) = 5.2 \times 10^{-3}$
Next step is to calculate the density of states associated with the conduction electrons of a metal from equation
$N(E) = CE $
$C = \frac{8\sqrt 2 \pi (9.109\times 10^{-31} kg)^{3/2}}{(6.626\times 10^{-34} J.s)^3}$
$C = 1.062 \times 10^{56} kg^{3/2}/J^3.s^3$
So,
$N(E) = (1.062 \times 10^{56} kg^{3/2}/J^3.s^3)(4.00 \times 10^{-19}J) $
$N(E) = 6.717 \times 10^{46} /m^3.J$
(The unit has converted since $1 kg = 1 \space J.s^2.m^{-2}$ )
Then, find the number of occupied state using the formula
$N_0(E) = P(E)N(E) $
$N_0(E) = (6.717 \times 10^{46} /m^3.J) (5.2 \times 10^{-3})$
$N_0(E) = 3.49 \times 10^{44} /m^3.J$
Lastlly,
Given the range of energy $\Delta E = 3.20 \times 10^{-20}J$
and volume, $V = 6.00 \times 10^{-6} m^3 $
$[Number \space of \space Occupied \space States] = N_0(E) V \Delta E$
$[Number \space of \space Occupied \space States] = (3.49 \times 10^{44} /m^3.J) ( 6.00 \times 10^{-6} m^3) (3.20 \times 10^{-20}J)$
$[Number \space of \space Occupied \space States] = 6.7 \times 10^{19}$
