(c)

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 $\geq$ weight of balloon and its payload

$\Rightarrow$ (Volume of air displaced $\times$ density of atmospheric air $\times$ Acceleration due to gravity) $\geq$ (Volume of air of inside balloon $\times$ density of air inside balloon $x$ acceleration due to gravity) $+$ (Weight of payload of balloon)

$\Rightarrow V \cdot \rho_o \cdot g \geq V \cdot \rho_i \cdot g+210 \times g$

where $\rho_0=$ density of outside air and $\rho_i=$ density of inside air.

$\Rightarrow \quad V\left(\rho_o-\rho_i\right)=210$

$\Rightarrow \quad \rho_0-\rho_i=\frac{210 \times 3}{4 \pi r^3}\left(\because V=\frac{4}{3} \pi r^3\right)$

$\Rightarrow \frac{P M}{R T_o}-\frac{P M}{R T_i}=\frac{210 \times 3}{4 \times \pi \times\left(\frac{11.7}{2}\right)^3} \Rightarrow \frac{1}{T_o}-\frac{1}{T_i}$

$=\frac{680\times 8 \times 8.31}{4\times \pi \times(1.7)^3 \times 10^5 \times 30 \times 10^{-3}}$

$\Rightarrow \frac{T_i-T_o}{T_o T_i}=\frac{1}{1387}$

$\Rightarrow T_o T_i=1387\left(T_i-T_o\right)$

$\Rightarrow 300 T_i=1387 T_i-300 \times 1387$ (as, $T_o=27^{\circ} C =300 K$ )

So, $T_i=\frac{300 \times 1387}{1087} \approx 383\, K$

$\therefore T_i=383-273=110^{\circ} C$

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