Initial velocity of the rocket, $v=2 km / s =2 \times 10^{3} m / s$

Mass of Mars, $M=6.4 \times 10^{23} kg$

Radius of Mars, $R=3395 km =3.395 \times 10^{6} m$

Universal gravitational constant, $G=6.67 \times 10^{-11} N m ^{2} kg ^{-2}$

Mass of the rocket $=m$

Initial kinetic energy of the rocket $=\frac{1}{2} m v^{2}$

Initial potential energy of the rocket $=\frac{-G M m}{R}$

Total initial energy $=\frac{1}{2} m v^{2}-\frac{ G M m}{R}$

If $20 \%$ of initial kinetic energy is lost due to Martian atmospheric resistance, then only

$80 \%$ of its kinetic energy helps in reaching a height.

Total initial energy available $=\frac{80}{100} \times \frac{1}{2} m v^{2}-\frac{ GMm }{R}=0.4 mv ^{2}-\frac{ GMm }{R}$

Maximum height reached by the rocket $=h$

At this height, the velocity and hence, the kinetic energy of the rocket will become zero.

Total energy of the rocket at height $h$

At this height, the velocity and hence, the kinetic energy of the rocket will become zero.

Total energy of the rocket at height $h \quad=-\frac{ G M m}{(R+h)}$

Applying the law of conservation of energy for the rocket, we kan write

$0.4 m v^{2}-\frac{G M m}{R}=\frac{-G M m}{(R+h)}$

$0.4 v^{2}=\frac{G M}{R}-\frac{G M}{R+h}$

$\quad=G M\left(\frac{1}{R}-\frac{1}{R+h}\right)$

$\quad=G M\left(\frac{R+h-R}{R(R+h)}\right)$

$\quad=\frac{G M h}{R(R+h)}$

$\frac{R+h}{h}=\frac{G M}{0.4 v^{2} R}$

$\frac{R}{h}+1=\frac{G M}{0.4 v^{2} R}$

$\frac{R}{h}=\frac{G M}{0.4 v^{2} R}-1$

$h=\frac{R}{\frac{R M}{0.4 v^{2} R}-1}$

$=\frac{0.4 R^{2} v^{2}}{G M-0.4 v^{2} R}$

$=\frac{0.4 \times\left(3.395 \times 10^{6}\right)^{2} \times\left(2 \times 10^{3}\right)^{2}}{6.67 \times 10^{-11} \times 6.4 \times 10^{23}-0.4 \times\left(2 \times 10^{3}\right)^{2} \times\left(3.395 \times 10^{6}\right)}$

$=\frac{18.442 \times 10^{18}}{42.688 \times 10^{12}-5.432 \times 10^{12}}=\frac{18.442}{37.256} \times 10^{6}$

$=495 \times 10^{3} m =495 km$