Decrease in Kinetic energy = increase in gravitational potential energy of particle

$\large\frac{1}{2}$$ mv^2 =\large\frac{mgh}{1+\Large\frac{h}{R}}$

$\therefore h= \large\frac{v^2}{2g-\Large\frac{v^2}{R}}$

$\qquad=\large\frac{(4.0 \times 10 ^3)^2}{2 \times 9.8 -\Large\frac{(4.0 \times 10 ^3)^2}{6.4 \times 10^6}}$

$\qquad=9.35 \times 10 ^5 \;m$

we get $ h= 935 \;km$

Hence d is the correct answer.