= Solution
Treat the <gravitational field> as approximately spherical. The <circular orbit> condition gives
$$
\boxed{M(<8\,\mathrm{kpc})=\frac{v_c^2r}{G}=9.0\times10^{10}M_\odot,\qquad \Upsilon=\frac{M/M_\odot}{L/L_\odot}=4.50.}
$$
If every star had a representative mass $m$ and obeyed $L/L_\odot=(m/M_\odot)^3$, its <mass-to-light ratio> would be $(m/M_\odot)^{-2}$, giving $\boxed{m\simeq0.47M_\odot}$. This is a one-mass estimate, not an actual average inferred uniquely from the data: a <galactic stellar mass function>, stellar remnants, dust and nonstellar mass alter the relation. The flattening of the <galactic disk> also changes the spherical conversion by an order-one factor.
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