Answer
The device is providing energy to the circuit.
Work Step by Step
We can find the potential difference across $R_3$:
$V_3 = (6.0~A)(6.0~\Omega) = 36~V$
Then the potential difference across $R_1$ is: $~~78~V-36~V = 42~V$
We can find the current through $R_1$:
$i_1 = \frac{42~V}{2.0~\Omega} = 21~A$
Then the current through $R_2$ is: $~~21~A-6.0~A = 15~A$
We can find the potential difference across $R_2$:
$V_2 = (15~A)(4.0~\Omega) = 60~\Omega$
Since the total potential difference across $R_2$ and the device must decrease by $36~V$, the potential must increase by $24~V$ across the box.
Therefore, the device must be providing energy to the circuit.
