If you model the noise of the resistor R between Gate and Drain of the transistor in Fig 3. a) by a noisless resitor with a noisy current source in parallel you can break up this current source in two current sources, one connected at the gate and one on the source. For an examplanation why this is possible, take a look at the figure A.1-8 in appendix A of Allen,Holberg CMOS Analog Circuit Design 2nd edition.

Now, the current source connected to the drain is in parallel with the transitor and so the current source noise contribution is canceled in the same way as the noise contribution of the transistor. The other current soure connected to the gate is connected in parallel to the voltage source Vs which can be transformed to a equivalent current source also. So the noise conrtibution of this current source is magnified in the same way as the input signal.

Hence the noise of the feedback resistor R cannot be canceled.