Sorry it took me so long. I couldn't find my book.
First, there is probably some way to fit this into a neat formula. But I didn't work this using these Norton equations. A lot of times it is hard to fit these problems exactly into a given equation so I used common sense and a little circuit theory.
Second, I hate Siemens. I alway convert to ohms.
First, the Norton current is equal to the short circuit current, or 3A. Even if you didn't know this, assume the network is a resistor with a current source. Connect a short across the resistor and obviously all the current from the source flows through the short, or 3A. So, you now have either answer B or D.
Draw the picture. You have a 3A source with two resistors across it. One is unknown and the other is .2 S or 1/.2 = 5 ohms.
Since the voltage across the parallel resistors is going to be 5V, that means (by ohms law) the equivalent resistance will be R = V/I = 5/3 = 1.67 ohm.
You have two choices for the unknown --> 1/.4 = 2.5 ohms, or 1/.6 = 1.67 ohm. So, which resistance, in parallel with the given 5 ohms, will give 1.67 ohm. You don't even have to calculate. It can't be the 1.67 ohm, it has to be 2.5.
But just in case, try 2.5 --> (2.5*5)/(2.5+5) = 12.5/7.5 = 1.67 (this is the parallel combined value I want). So I pick B = 1/2.5 ohm = .4 S.
Hopefully you can understand my explanation, or somebody will explain the plug in method, but this is about as tough of a circuits problem as you'll get in the AM. They can have little tricks, and drawing the picture and doing a little thinking pays off.