Answer
$7.78\times10^{17}kg$
Work Step by Step
Please see the attached image first.
Let's take,
mass of the projectile = M
Radius of the mars = $3.39\times10^{6}m$
Mass of the Mars = $0.642\times10^{24}kg$
In this case, there is no external torque. So its angular momentum is conserved.
We can write,
$I_{1}\omega_{1}=I_{2}\omega_{2}$ ; Let's plug known values into this equation.
$\frac{2}{5}\times0.624\times10^{24}kg\times(3.39\times10^{6}m)^{2}\times\frac{1}{1.03\times24\times3600\space s}=\frac{2}{5}\times0.624\times10^{24}kg\times(3.39\times10^{6}m)^{2}\times\frac{1}{1\times24\times3600\space s}-M\times3.39\times10^{6}m\times 2.44\times10^{6}m/s$
$8.27\times10^{12}=\frac{18.54\times10^{36}}{3600\times24}(1-\frac{1}{1.03})kg$
$8.27M=2.15\times10^{20}(0.03)$
$M=7.78\times10^{17}kg$
