How to work out how much pulling power an electric motor has in kg?
Mechanical power required, in watts, is \$T\omega\$, where \$T\$ is torque in \$Nm\$ and \$\omega\$ is rotational speed in \$rad\:s^{-1}\$. You might need to double this power to arrive at the electrical motor power required in order to overcome friction losses in the gearbox or linkage and also to take into account motor losses.

In your case the mechanical power required is about \$3\:W\$.