Answer
$a\approx1000~g$ (other approximations are also possible)
Work Step by Step
The front surface of the bag expands from rest ($v_0=0$) with constant acceleration, $a$. Suppose that the front surface of the bag expands $50~cm=0.5~m$.
$g\approx10~m/s^2$
$x=x_0+v_0t+\frac{1}{2}at^2$. Now, assuming that $x_0$ is the origin:
$0.5~m=0+0(10\times10^{-3}~s)+\frac{1}{2}a(10\times10^{-3}~s)^2$
$0.5~m=0.5a(10^{-2}~s)^2$
$0.5~m=(0.5\times10^{-4}~s^2)a$
$a=\frac{0.5~m}{0.5\times10^{-4}~s^2}=10^{4}~m/s^2$
$a=(10^{4}~m/s^2)(\frac{1~g}{10~m/s^2})=1000~g$
