Answer
$2.919\times 10^{-9}\;\rm e/atom$
Work Step by Step
We know that the number of charge carriers per unit volume in a current is given
$$n=\dfrac{I}{ev_{\rm d}A}\tag 1$$
whereas $I$ is the current, $e$ is the charge of one electron, $A$ is the cross-sectional area of the silicon strip, and $v_{\rm d}$ is the drift velocity which is given by
$$v_{\rm d}=\dfrac{V_{\rm H}}{Bd}$$
whereas $V_{\rm H}$ is the hall effect voltage, $B$ is the magnetic field strength, and $d$ is the width of the silicon strip.
Plugging into (1);
$$n=\dfrac{IBd}{eV_{\rm H}A} $$
Noting that the cross-sectional area of the silicon strip is given by $A=td $, and hence, $t=A/d$.
So that
$$n=\dfrac{IB}{eV_{\rm H}t} $$
Plugging the given;
$$n=\dfrac{(0.28\times10^{-3})(1.5)}{(1.6\times 10^{-19}) (18\times 10^{-3})(1\times 10^{-3})} $$
$$n= \bf 1.458\times 10^{20}\;\rm e/m^3$$
Now we need to find the number of silicon atoms in one cubic meter.
$$N=\dfrac{\rho}{m}\times N_A$$
where $m$ is the mass of one mole of silicon, and $N_A$ is Avogadro's number which is the number of atoms in one mole.
$$N=\dfrac{2330}{28.0855 \times 10^{-3}}\times 6.02\times 10^{23}$$
$$N=\bf 4.99425\times 10^{28}\;\rm atom/m^3$$
Therefore, the number of electrons per silicon atom which are in the conduction band is given by
$$N_e=\dfrac{n}{N}$$
Plugging from above;
$$N_e=\dfrac{1.458\times 10^{20}}{4.99425\times 10^{28}}$$
$$N_2=\color{red}{\bf 2.919\times 10^{-9}}\;\rm e/atom$$