b
(B)
For equilibrium :
$mg ^{\prime}=\frac{1}{2} \rho v ^2 A$
$mg \left(1-\frac{2 h }{ R }\right)=\frac{1}{2} \times 1.2 e ^{-\frac{ h }{10}} v ^2 A$
$\Rightarrow 7500 \times 9.8\left(1-\frac{2 h }{ R }\right)=0.6 e ^{-\frac{ h }{10}}\left(8 \times 10^3\right)^2 \times 30$
$\downarrow$ Neglect this term for a while and we get $h \simeq 95 \,km$
So, with $\left(1-\frac{2 h}{R}\right)$, we will get little lesser value or $h$.
$\therefore(B)$ is the answer.
