= Solution
\b[The extra radial term.] A <circular Kepler orbit> has $\Omega_{\rm orb}^2=GM/R^3$. In the frame rotating with the synchronously spinning body, centrifugal acceleration increases by $\Omega_{\rm orb}^2a$ across a radial offset $a$. This adds the outward differential acceleration
$$
\boxed{\Delta a_{\rm cent}=\frac{GM}{R^3}a.}
$$
Equivalently, in the inertial frame the farther corotating point needs an extra inward centripetal acceleration $\Omega_{\rm orb}^2a$, which gravity alone does not supply. The total radial stretching relevant to the static corotating body is therefore $3GMa/R^3$. This is the <corotating point-mass tidal field>. The source's “force” expression is per unit mass.
\b[Self-gravitational survival.] Denote the asteroid radius by $s$, distinct from the orbital radius $R$. Its mass is $m_a=4\pi\rho s^3/3$ and its surface self-gravitational acceleration is $Gm_a/s^2=4\pi G\rho s/3$. Compare this restoring acceleration with radial tidal stretching $3GMs/R^3$:
$$
\frac{4\pi G\rho s}{3}\gtrsim\frac{3GMs}{R^3}.
$$
The <corotating tidal-disruption estimate for a uniform sphere> is
$$
\boxed{R_{\min}\simeq\left(\frac{9M}{4\pi\rho}\right)^{1/3}.}
$$
This is the requested spherical local force-balance estimate, not an exact fluid Roche figure. It neglects deformation and material cohesion.
\b[Why the asteroid radius cancels.] At fixed density, self-gravitational surface acceleration scales as $m_a/s^2\propto s$. The linear tidal acceleration also scales as $s$. Their ratio is therefore independent of $s$, provided $s\ll R$ so that the tidal expansion is valid. Density, rather than asteroid size, controls this estimate.
\b[The terrestrial value.] Convert $\rho=2\ \mathrm{g\,cm^{-3}}$ to $2000\ \mathrm{kg\,m^{-3}}$ and use $M_E=6.0\times10^{24}\ \mathrm{kg}$:
$$
\boxed{R_{\min}\simeq
\left[\frac{9(6.0\times10^{24})}{4\pi(2000)}\right]^{1/3}
\simeq1.29\times10^7\ \mathrm m
=1.29\times10^4\ \mathrm{km}.}
$$
This is distance from Earth's centre, not height above its surface.
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