How to add a particular noise on a signal?
Please see this post that details most of the noise considerations for an ADC

What are advantages of having higher sampling rate of a signal?

In particular the relationship for the noise floor given a full scale sine wave:

$$SNR = 6.02 \text{ dB/bit} + 1.76 \text{ dB}$$

Using your numbers:

1.5Vpp sinewave. If we assume this sinewave is full scale (any higher voltage would start to add clipping noise), then this is the level of 0 dBFS. In particular the variance of this or the standard deviation is used in the dB ratio computation according to:

$$SNR = 20Log_{10}(\sigma_s/\sigma_n)$$

or

$$SNR = 10Log_{10}(\sigma_s^2/\sigma_n^2)$$

Since the noise itself is given in units of dBFS, there is nothing to do further with the 1.5V level. This level is important to determine if the signal is indeed at full scale, and if we wanted to determine the noise level in units of volts.

120 dBFs/Hz with a 200 MHz Sampling Rate:

The quantization noise can be reasonably assumed to be a uniformly distributed white noise (see link above). So the total noise in 200 MHz of bandwidth would be:

$$-120 \text{ dBFS} + 10Log_{10}(200e6) = -37.0 \text{ dB}$$

(Note that this is the double-sided noise in the spectrum extending from $-f_s/2$ to $+f_s/2$)  

From the formula above we see that this would be 6.16 bits of precision. A 6 bit converter with the sinewave right at full scale would be very close the the added noise that was given in the example.

To "add" such noise to the signal, scale the signal to fit peak-peak with a 6 bit system and round to closest integer to create the resulting quantization noise.

So the +/-1.5V peak to peak sine wave would then be scaled to $0$ to $2^6$, or $-2^5$ to $+2^5-1$ if you want to represent a signed signal.  The noise will occur from either rounding or truncating to integer values. (There is no noise difference in rounding or truncation; truncation simply introduces an offset).

As far as the FFT, each bin in the FFT if not windowed has a bandwidth 1 bin wide. So the FFT noise floor will go down according to the number of bins. This is explained further in this post:

How large FFTs can pull signals out of the noise floor?