Constructing a dc circuit with different input and output currents and voltages?
(i) What I understand from the circuit is that you have to replace the box in figure with an equivalent circuit. 

If Vin is the applied voltage and the current is Iin, we can represent the input port with a resistance whose value is Rin = Vin/Iin. But since Iin is too small, the value of Rin will be very high.

The circuit at output is already given in the question. It is a voltage source and a series resistance. So the very simple circuit that can replace the box given in question is 







simulate this circuit &ndash; Schematic created using CircuitLab

(ii) When a subsequent resistor R of is connected at the output as shown in figure below,
then






simulate this circuit

$$I_{out} = \frac{V_{out}}{R+r}\tag1$$

and if \$r &gt;&gt; R\$, then we can neglect R and write \$R+r \approx r\$ then 
$$I_{out} = \frac{V_{out}}{r}$$ 

ie, the output current is independent of the resistor R and the output act as a constant current source, provided the internal resistance r of the circuit is higher than R.

Writing the expression for voltage across R,
$$V_R = I_{out}\times R$$ 
from (1),
$$ V_R = \frac{V_{out}}{R+r} \times R$$
$$= V_{out} \times \frac{1}{1 + r/R}$$

If this should work as a voltage source, then \$V_R\$ should be equal to \$V_{out}\$. Or,
$$ \frac{1}{1 + r/R} = 1$$
$$\Rightarrow r/R &lt;&lt; 1$$
Or,
$$r &lt;&lt; R$$
ie, the internal resistance should be less than the subsequent resistance.

Read about internal resistance of a current source and voltage source.