Answer
$50~kg~~$ of gas must be ejected each second.
Work Step by Step
We can find $R$, the rate at which the gas must be ejected:
$R~v_{rel} = F_{thrust}$
$R = \frac{F_{thrust}}{v_{rel}}$
$R = \frac{mg}{v_{rel}}$
$R = \frac{(6100~kg)(9.8~m/s^2)}{1200~m/s}$
$R = 50~kg/s$
$50~kg~~$ of gas must be ejected each second.
