How to increase current of a DC circuit
You can't. Actually it's all Ohm's fault:  

\$ I = \dfrac{V}{R} \$

The current is defined by voltage and resistor. If you have 2 mA at 5 V then your resistor will be 2500 &Omega;. End of story.  

If you want to increase the current you'll have to 1)increase your voltage, 2)decrease your resistance, or 3) a combination of both. For instance:

\$ I = \dfrac{V}{R} = \dfrac{50 V}{250 \Omega} = 200 mA \$

or also   

\$ I = \dfrac{V}{R} = \dfrac{5 V}{25 \Omega} = 200 mA \$

A transistor can be used to increase current. You'll have a low current path, from base to emitter in an NPN, and a higher current path from collector to emitter. The collector current will be a multiple of the base current if the circuit allows it. That means that the voltage source at the collector side must be high enough, and the load resistance low enough to get the 200 mA. Again Ohm's Law: if you have a 1 k&Omega; collector resistor you will never get more than 5 mA with a 5 V supply, no matter how hard the transistor will try to draw more.



About your comment on controlling a LED. First 200 mA is a lot for a common LED, too much actually. A value like 20 mA will be more than enough for an indicator LED. Your LED will have about 2 V across it, that's a voltage you subtract from the 5 V. The remaining 3 V will be across the series resistor. So we apply Ohm, again:

\$ R = \dfrac{V}{I} = \dfrac{3 V}{0.02 A} = 150 \Omega; \$

If you use a 150 &Omega; resistor in series with the LED you'll get 20 mA. You don't have to amplify anything for that; it's the resistance which dictates the current. If you would simply use the 150 &Omega; resistor without the LED then the full 5 V will be across the resistor and the current will be 33 mA = 5 V / 150 &Omega;. Again no need for amplification, just changing the resistance or voltage will do.