= Solution
Let $V=4\pi R^3/3$, $w_p=w(Z)$ and $\rho_p=\rho(Z)$, with the subscript $p$ here referring to the plume fluid. The section at $Z(t)$ moves upward at $U=\dot Z$. Its incoming <volume flux> relative to that moving section is therefore
$$
J=\pi B^2(w_p-U).
$$
The thermal's sectional model allows no direct ambient <fluid entrainment>, so its volume and mass balances are
$$
\boxed{\dot V=J,\qquad \frac{d}{dt}(\rho_tV)=\rho_pJ.}
$$
Subtracting these balances with $g_t=g(\rho_0-\rho_t)/\rho_0$ and $g_p=g(\rho_0-\rho_p)/\rho_0$ gives the <buoyant thermal mass balance>
$$
\frac{d}{dt}(g_tV)=g_pJ=g_p\dot V.
$$
For the <self-similar starting plume>, $V\propto t^{9/4}$ and $g_t\propto t^{-5/4}$, so $g_tV\propto t$. Hence
$$
\frac{g_tV}{t}=g_p\frac94\frac Vt,\qquad
\boxed{g_t=\frac94g_p.}
$$
Using $\pi B^2w_p$ instead of the relative inflow would miss the volume swept out by the rising matching section. The larger thermal <reduced gravity> reflects fluid accumulated from earlier, lower, more buoyant parts of the plume.
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