= Solution
Assume a constant downward <gravitational acceleration> $g$, constant composition, and an <ideal gas law> $p=\rho\mathcal R T$, where $\mathcal R=k_B/(\mu m_u)$ is the gas constant per unit mass. Here $m_u$ is the <unified atomic mass unit>; this convention makes the specified <mean molecular weight> dimensionless. <Hydrostatic equilibrium> gives
$$
\frac{dp}{dz}=-\rho g=-\frac{pg}{\mathcal R(T_0-\alpha z)},\qquad
\frac{d\ln p}{dz}=-\frac{g}{\mathcal R(T_0-\alpha z)}.
$$
Integrating from sea level yields
$$
\frac{p(z)}{p(0)}=\left(\frac{T_0-\alpha z}{T_0}\right)^{g/(\mathcal R\alpha)}.
$$
Dividing by the <temperature> ratio gives the <hydrostatic atmosphere with a constant temperature lapse rate>:
$$
\boxed{\rho(z)=\rho(0)\left(1-\frac{\alpha z}{T_0}\right)^{\mu m_ug/(k_B\alpha)-1}.}
$$
The model requires $T_0-\alpha z>0$ and is used here only within the troposphere.
Convert the <atmospheric lapse rate> to $\alpha=6.5\times10^{-3}\ \mathrm{K\,m^{-1}}$. At $11\ \mathrm{km}$, $T=228.5\ \mathrm K$. With $g=9.81\ \mathrm{m\,s^{-2}}$, $k_B=1.380649\times10^{-23}\ \mathrm{J\,K^{-1}}$, $m_u\simeq1.66054\times10^{-27}\ \mathrm{kg}$ and $\mu=28$, the density exponent is $4.0825$. Hence
$$
\boxed{\frac{\rho(11\ \mathrm{km})}{\rho(0)}
=\left(\frac{228.5}{300}\right)^{4.0825}\simeq0.329.}
$$
The <mass density> is therefore approximately one third of its sea-level value in this model.
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