Hot air balloon will rise in the atmosphere when upthrust of buoyant force is greater than weight of balloon and its payload. Upthrust = Weight of atmospheric air displaced by balloon So, upthrust $\ge$ weight of balloon and its payload

$\Rightarrow$ (Volume of air displaced $\times$ density of atmospheric air $\times$Acceleration due to gravity) $\ge$ (Volume of air of inside balloon

x density of air inside balloon $\times$ acceleration due to gravity) + (Weight of payload of balloon) $\Rightarrow V·{\rho }_o·g\ge V·{\rho }_i·g+210\times g$

where ${\rho }_o=$ density of outside air and ${\rho }_i=$ density of inside air.

$\Rightarrow V\left({\rho }_o-{\rho }_i\right)=210$ $\Rightarrow {\rho }_0-{\rho }_i=\frac{210\times 3}{4\pi r^3}\left(\because V=\frac{4}{3}\pi r^3\right)$ $\Rightarrow \frac{PM}{R{T}_0}-\frac{PM}{R{T}_i}=\frac{210\times 3}{4\times \pi \times {\left(\frac{11.7}{2}\right)}^3}\Rightarrow \frac{1}{T_0}-\frac{1}{T_i}$ $=\frac{680\times 8\times 8.31}{4\times \pi \times (11.7{)}^3\times {10}^5\times 30\times {10}^{-3}}$ $\Rightarrow \frac{T_i-T_0}{T_0{T}_i}=\frac{1}{1387}$

$\Rightarrow T_0{T}_i=1387\left(T_i-T_o\right)$ $\Rightarrow 300T_i=1387T_i-300\times 1387$ $\begin{array}{l}\\ \text{ (as, }\left.T_0={27}^{\circ }\mathrm{C}=300\mathrm{K}\right)\end{array}$ $\text{ So, }{T}_i=\frac{300\times 1387}{1087}\approx 383\mathrm{K}$ $\therefore T_i=383-273={110}^{\circ }\mathrm{C}$

So, temperature of hot air is near to $105°\mathrm{C}$.