= Solution
For the same ideal <gas> composition and temperature, $\rho=pM/(RT)$, hence $\Delta\rho/\rho_0=\Delta p/p_0=10^{-5}$. The bubble volume is $V=4\pi r^3/3=4.189\times10^{-3}\,\mathrm{m^3}$. The <gas>'s mass in excess of displaced ambient air is
$$
\boxed{m_b=\Delta\rho V=(1.2\times10^{-5})(4.189\times10^{-3})=5.03\times10^{-8}\,\mathrm{kg}=50.3\,\mu\mathrm g.}
$$
This is a \b[positive buoyant mass], meaning a net downward weight after subtracting ordinary air buoyancy. It is not an upward lifting capacity. Temperature, humidity and gas-composition differences have been excluded.
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