Initially, both the blocks were under equilibrium, so for mass $m$,
$T=mg$, 
and for mass $2m$,
$F_{\mathrm{s}}=T+2mg=3mg$
Now, after the string is cut, the tension in it becomes zero. Therefore, 
for $m$,
$mg-0=ma \Rightarrow a=g$(downward)
and for $2m$,
$$\begin{gathered} 2mg+0-F_{\mathrm{s}}=2mb \\ 2mg+0-3mg=2mb \\ \Rightarrow b=-\frac{g}{2} \end{gathered}$$
i.e., $\frac{g}{2}$ upward.