Answer
$11.8\space km/s$
Work Step by Step
Here we use the principle of the conservation of mechanical energy.
Final mechanical energy = Initial mechanical energy
$\frac{1}{2}mV_{S}^{2}- \frac{GM_{E}m}{r_{m}}=\frac{1}{2}mV_{0}^{2}-\frac{GM_{E}m}{R_{E}}$
$V_{S}^{2}=V_{0}^{2}+2GM_{E}(\frac{1}{r_{m}}-\frac{1}{R_{E}})$
Where, $V_{S}$ - speed at the moon's orbit, $V_{0}$ - Initial speed, $M_{E}$ - mass of the earth, $r_{m}$- distance to moon's orbit from earth's center, $R_{E}$- radius of the earth.
Let's plug known values into the above equation.
$V_{S}^{2}= (16.26\times10^{3}m/s)^{2}+2\times6.67\times10^{-11}Nm^{2}/kg^{2}\times5.97\times10^{24}kg(\frac{1}{3.9\times10^{8}m}-\frac{1}{6.37\times10^{6}m})$
$V_{S}^{2}=264.4\times10^{6}\space m^{2}/s^{2}-125.3\times10^{6}\space m^{2}/s^{2}$
$V_{S}^{2}=139.1\times10^{6}m^{2}/s^{2}$
$V_{S}= 11.8\times10^{3}m/s=11.8\space km/s$
