Answer
$\rm 46^\circ C$
Work Step by Step
First of all, we need to draw the force diagram of the balloon and its equipment, as we see in the figure below.
We know that the balloon will move up at a constant velocity which means that the net force exerted on it is zero.
$$\sum F_y=F_B-W_{b}-W_{g}=M_{tot}a_y=M_{tot}(0)=0$$
whereas $F_B$ is the buoyant force which is known as the weight of the cold air displaced by the volume of the balloon $(W_{cold})$, $W_b$ is the weight of the balloon and its equipment, and $W_{hot}$ is the weight of the gas inside the balloon.
Thus,
$$F_B=W_{b}+W_{hot}$$
$$W_{cold}=W_{b}+W_{hot}$$
$$m_{cold}g=3300+m_{hot}g$$
Recall the density law of $\rho =m/V$, and hence $m=\rho V$
$$\rho_{cold}Vg=3300+ \rho_{hot}Vg\tag 1$$
We know that the law of ideal gas is given by $PV=nRT$
whereas $n=\dfrac{m}{M}$ where $m$ is the mass of the gas and $M$ is its molecular mass.
Thus,
$$PV=\dfrac{m}{M}RT$$
$$PM= \overbrace{ \dfrac{m}{V}}^{{\color{red}=\;\rho}}\;RT $$
Hence,
$$\dfrac{PM}{R}=\rho T\tag 2$$
Now we know that the gas inside and outside the balloon is the air which means that the molecular mass of both is the same. Also, the pressure inside the balloon is equal to the atmospheric pressure since the balloon is not isolated from the surrounding environment.
Thus, the left side in equation (2) is constant (recall that $R$ is also a universal constant).
Therefore,
$$\rho_{cold} T_{cold}=\rho_{hot}T_{hot}$$
Solving for $\rho_{hot}$;
$$\rho_{hot}=\dfrac{\rho_{cold} T_{cold}}{T_{hot}}$$
Plug into (1);
$$\rho_{cold}Vg=3300+ \dfrac{\rho_{cold} T_{cold}}{T_{hot}}Vg$$
Solving for $T_{hot}$;
$$\rho_{cold}Vg-3300= \dfrac{\rho_{cold} T_{cold}}{T_{hot}}Vg$$
$$T_{hot}= \dfrac{\rho_{cold} T_{cold}}{ \rho_{cold}Vg-3300 }Vg$$
Plugging the known;
$$T_{hot}= \dfrac{1.29\cdot 273 }{[1.29 \cdot 1800\cdot 9.8]-3300 }\cdot 1800\cdot 9.8$$
$$T_{hot}=319 \; \rm K ={\bf 46}^\circ C$$
The factor that limits the maximum altitude attainable for this case is the change in the air's density as we go up.
The outer air became thinner and thinner which means that its density decreases so much, and to make the inner air less dense than the outer air then we need to raise the temperature of the inner air to some degrees that may be too dangerous.