$KE$ given =$P.E$ of the body
$\large\frac{1}{2} $$mv^2=mg (R+h)$
$v^2=2g(R+h)$
Where $'h'$ maximum height reached by the satellite
$\bigg( 10 \times \large\frac{5}{8}\bigg)^2 $$\times 10^6= 2\times 10(R+h)$
Solving we get
$R+h=5R$
Where $R= 6.4 \times 10^6 \;m$
$R+h=5R$
$\quad h=4R$
Hence b is the correct answer.