Finding delay parameters for watermarked signal
Thank you for posting the additional information, it certainly helped.  I'll try to do a prose explanation with less "math" in it, and then point you in what is the right direction.

That's certainly a long winded problem statement, and I believe the solution is actually simpler that your professor may be making it out to be.  From reading the problem statement, a watermark has been made and was supplied to you.  This watermark is just sampled Gaussian noise, such that it lacks coherent frequency content, i.e. the spectral energy of the watermark is not located at any specific frequency (as opposed to say, a constant frequency signal like a sinusoid).

So we've got our original signal, we have our watermark signal, now it's time to actually watermark.  The simplest approach would just be adding the watermark signal to the original signal.  Your professor has gone a step ahead and added 5 copies of the signal.  Each one of these copies has its own time delay, as well as its own amplitude.  So, after we  add in our 5 watermarks to our signal, we have the signal Xm

Since there are 5 copies of the signal at 5 unique points in time in Xm, that means if we did a cross-correlation with our watermark signal and Xm, we would expect to see 5 distinct peaks at different amplitudes. This operation is typically called a matched filter, and you can think of it as using a finding the "locations" (in your professor's words, the delay parameters; think of this as when each watermark "starts" in Wm) of some given template (i.e. the watermark) in your recorded signal.

Your professor has gone ahead and detailed out how you can accomplish cross-correlation in the frequency domain.  This is often the preferred method, since it is typically much more computationally efficient.  Your professor has also detailed common ways to improve cross-correlation detection, specifically the use of windowing functions as well as the use of frequency normalization.  

In turns out  those methods he described will be very important for your success.  The watermark signals are pretty low gain (naturally), so we'll need to maximize the signal to noise ratio (SNR) as much as possible.

I think that since this is seemingly an assignment of some sort, I'm not willing to just give you the answer in numeric or code form, but here's a pictorial output of what I got:


You can see that I've marked the 5 points corresponding to the time delayed version of the signal with red X marks.  These  peaks are easily 5x larger than any other point in the output of the cross-correlation, so one can easily  conclude that these are statistically significant, and point to the location of the delayed watermark signals.

Here's some code I used.  Feel free to play around with it and see if you get satisfactory results (note: "data" is your Xm signal, and "wm" is your watermark signal:

% Determine number of samples for the correlation lag vector
nsamps = numel(data);

% define a window function. Taylor windows give great performance
win    = taylorwin(nsamps,7,-10);

% window each signal
xin    = data;
wmw    = win .* wm;

% make the signals complex for phase coherency
xin    = hilbert(data);
wmw    = hilbert(wmw);

% perform cross-correlation in the frequency domain
xf     = ( fft((xin)) .* conj(fft((wmw))) );
% transform out of the frequency domain while normalizing
x      = ifft(xf ./ (abs(xf)+1));


lags   = linspace(-nsamps/2, nsamps/2, nsamps);

figure(1); hold off;
plot(lags,fftshift(abs(x)))
% find the 5 peaks, then plot them
[pks,locs] = findpeaks(fftshift(abs(x)),'NPeaks',5,'MinPeakHeight',0.0075);
hold on;
plot(lags(locs),pks,'rx');
xlabel('Correlation Lags'); ylabel('Scaled Correlation Magnitude (dB)')
grid on;
title('Watermark Cross-Correlation');


So hopefully that gets you going in the right direction.  Like I mentioned before, the watermark signals are pretty low gain, so make sure you do that normalization.  The windowing function is arguably not necessary, but they're pretty standard in practice since  they help your cross-correlation "behave" a bit better.