Answer
$v=5.92\times10^3m/s$
Work Step by Step
We have the period of the satellite $T=1.2\times10^4s$. We also know $M_E=5.98\times10^{24}kg$ and $G=6.67\times10^{-11}Nm^2/kg^2$
First, we try to find the radius of the orbital path $r$ by
$$T=\frac{2\pi r^{3/2}}{\sqrt{GM_E}}$$ $$r^{3/2}=\frac{T\sqrt{GM_E}}{2\pi}=3.81\times10^{10}m^{3/2}$$ $$r=1.13\times10^7m$$
Then, from here, we can find $v$ $$v=\frac{2\pi r}{T}=5.92\times10^3m/s$$
