The correct option is A
Let $\mathrm{F}_{\mathrm{b}}$ be the up thrust of air and then for downward motion,
$\mathrm{mg}-\mathrm{F}_{\mathrm{b}}=\mathrm{ma}$
Let $\mathrm{m}$’ be the mass removed from the balloon
So when it start moving upward, we have,
$\mathrm{F}_{\mathrm{b}}-\left(\mathrm{m}-\mathrm{m}^{\prime}\right)=\left(\mathrm{m}-\mathrm{m}^{\prime}\right) \mathrm{a}$
$\mathrm{m}^{\prime}=\frac{2 \mathrm{ma}}{\mathrm{g}+\mathrm{a}}$