How to estimate the transfer function given only Bode plot data points?
it is not that complicated, watch for the point the graph gets a slope downwards that is a pole. That is when the denominator of the transfer function is 0... later on that downward slope stops and becomes flat so there must be a 0 canceling out the pole earlier. This happens at 1kHz for the pole and at 1MHz for the 0.

You usually write bode functions in Laplace form so switch out the S for w*t where w is equal 2*pi*f and do not forget about the imaginary term.

t is time, pi is just pi, f is frequency you can solve for a given frequency.

now for the constant, your graph starts at 0 db and it is flat so it must be 1 normally you solve for how many Dbs it is at 0 hertz, in your case it starts at 0 so to summarize your bode plot looks like this

a constant A that is equal to the number of Decibels at 0 frequency since 20*log(A)=0 your frequency is 0 here so any S term is gone.

at 1Khz you have a pole so the denominator is 0 so s= 2*pi*f at 1kHz so  s+2000*pi is your pole now your function looks like this the extra minus comes from the imaginary term. 
$$
F(s)=\frac{A}{S+2000\pi}
$$
 you can rewrite this as 
$$
\frac{A}{(\frac{S}{2000\pi})+1}
$$
finally you have the remaining 0 at 1MHz so (S/1e6)+1

so the complete thing is F(s)= A times the pole times the zero 

F(s)= A*((S/pi*1e6)+1 )/((S/2000*pi)+1) A is 1 so you can just ignore it in this case.

I did not do it perfectly but I think it explains the gist of it.