Solve a circuit with mesh analysis
You have three meshes to work out, and one of them (the lower left) has a current assigned to it already, which is \$3\,mA\$, thus you don't have to mind the lower left mesh, which reduces your problem to 2 meshes only. 
For the lower right one we have, assuming its current to be \$I_2\$ and the upper one \$I_3\$:
 \begin{equation}
20 +  2000\cdot(0.003 - I_2) - 4000\cdot(I_2 - I_3) - 1000\cdot(I_2) = 0
\end{equation}
And for the last equation (upper mesh) yields:
\begin{equation}
4000\cdot(I_2-I_3) + 4000\cdot(0.003-I_3) - 1000\cdot(I_3) - 5 =0 
\end{equation}
We are left with 2 equations and 2 unknowns, now it's a winner winner chicken dinner!!

Solve for \$I_2\$ and \$I_3\$ to get the answers and then find \$V_o\$ from Ohm's law simply writing:

\$V_o = I_2\cdot1000\$

EDIT: Solving the equations above (currents are taken in clockwise direction) we have the outcome for \$ V_o\$ and \$ I_2\$:

\$ I_2 = 5.57 mA \$

\$ V_o = 1000\cdot I_2 = 5.57\,V\$