Answer
$2\;\rm nC,\;4\;nC$
Work Step by Step
Initially, the two spheres were not connected. So the net charge of the system is the charge on sphere 1.
$$Q_{i}=q_{i1}=6\;\rm nC$$
The two spheres are now connected, and the charge is conserved. So after enough time of exchanging charges between the two spheres, the net charge is still equal to $Q_i$ but is now divided on both spheres.
Thus,
$$Q_f=Q_i=q_{1f}+q_{2f}$$
Thus,
$$ q_{1f}+q_{2f}=6\;\rm nC\tag 1$$
Now we have one big conductor that consists of two spheres plus one thin wire that connects them. The whole system now must have the same electric potential since it is now in an electrostatic equilibrium.
Thus,
$$V_1=V_2$$
Hence,
$$\dfrac{ \color{red}{\bf\not} k_e q_{1f}}{R_1}=\dfrac{ \color{red}{\bf\not} k_e q_{2f}}{R_2}$$
We know that $R_2=2R_1$, so
$$\dfrac{ q_{1f}}{ \color{red}{\bf\not} R_1}=\dfrac{ q_{2f}}{2 \color{red}{\bf\not} R_1}$$
So,
$$q_{2f}=2q_{1f}\tag 2$$
Plug into (1),
$$ q_{1f}+2q_{1f}=3q_{1f}=6\;\rm nC $$
$$q_{1f}=\color{red}{\bf 2.0}\;\rm nC$$
Plug into (1), and solve for $q_{2f}$
$$q_{2f}=\color{red}{\bf 4.0}\;\rm nC$$
