= Solution
Take $z$ positive upward from the release and define the pressure scale height $H_p=p_0/(\rho_0g)$. Neglecting the small ambient density variation in the hydrostatic pressure calculation,
$$
p(z)=p_0-\rho_0gz=p_0(1-z/H_p).
$$
Conservation of gas mass and <Boyle's law> give
$$
\boxed{\rho_b(z)=\rho_B(1-z/H_p),\qquad
V_b(z)=\frac{\phi_0V_0}{1-z/H_p}.}
$$
Here $V_b(0)\simeq\phi_0V_0$ to leading order in the dilute bubble concentration. If the thermal's total volume is $V(z)$, its bubble fraction and bubble contribution to <reduced gravity> are
$$
\boxed{\phi(z)=\frac{\phi_0V_0}{V(z)(1-z/H_p)},\qquad g'_{\mathrm{bubbles}}\simeq g\phi(z).}
$$
Thus the volume-integrated kinematic buoyancy contribution is $g\phi V=g\phi_0V_0/(1-z/H_p)$, and the corresponding upward force is approximately $\rho_0$ times this quantity. Bubble expansion increases this total contribution even while entrainment dilutes the local concentration.
Neglect bubble rise relative to the water when the bubbles' terminal slip speed is much smaller than the thermal rise speed, their response time is short compared with $a/W$, and they remain well dispersed and retained. Small bubbles are more likely to satisfy these conditions; large bubbles can escape through the thermal. The approximations also require $\rho_b\ll\rho_0$, $\phi\ll1$, negligible gas dissolution and no important surface-tension correction. Pressure must remain positive and the hydrostatic expression is used only below the ocean surface.
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