Answer
$3.826\;m/s^2$
Work Step by Step
The ratio of the air density outside the balloon to that inside is $1.39$
$\frac{\rho_{out}}{\rho_{in}}=1.39$
The ratio implies that the the density of the air inside the balloon is less dense than the air outside the balloon. Therefore, the balloon experiences an upward buoyant force $(F_b)$. Simultaneously the weight $(W)$ of the balloon acts on vertically downward direction.
$F_b=\rho_{out}Vg$ and $W=\rho_{in}Vg$
where, $V$ is the volume of the balloon.
The resultant upward force acting on the balloon is
$F=F_b-W$
or, $\rho_{in}Va=\rho_{out}Vg-\rho_{in}Vg$
where, $a$ is the upward acceleration of the balloon.
or, $\rho_{in}a=\rho_{out}g-\rho_{in}g$
or, $a=(\frac{\rho_{out}}{\rho_{in}}-1)g$
Substituting the given values, we obtain
$a=(1.39-1)\times9.81\; m/s^2$
or, $\boxed{a=3.826\;m/s^2}$
Therefore, the acceleration of the rising hot-air balloon is $3.826\;m/s^2$
