a
\(\begin{array}{l}
\,\,\,\,\,\,\,\,Let\,F\,be\,the\,upthrust\,of\,the\,air,\,As\,the\,ballon\\
is\,descending\,down\,with\,an\,acceleration\,a,\\
\therefore \,\,mg - F = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
Let\,mass\,{m_0}\,be\,removed\,from\,the\,balloon\,so\\
that\,it\,starts\,moving\,up\,with\,an\,acceleration\,\\
a.\,\,Then,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F - \left( {m - {m_0}} \right)g = \left( {m - {m_0}} \right)a
\end{array}\)

\(\begin{array}{l}
\,\,\,\,\,\,\,F - mg + {m_0}g = ma - {m_0}a\,\,\,\,\,\,\,...\left( {ii} \right)\\
Adding\,eqn.\,\left( i \right)\,and\,eqn.\,\left( {ii} \right),\,we\,get\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_0}g = 2ma - {m_0}a\,;\,{m_0}g + {m_0}a = 2ma\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_0}\left( {g + a} \right) = 2ma\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_0} = \frac{{2ma}}{{a + g}}
\end{array}\)