We have an open loop system with an input u , a transfer function G(s) and an output y. We apply the following inputs ui(t)=sin(ωit),i=1,2,...6

These are of the form y(t)=Y1sin(ωit+ψi)
We have G(s)U(s)=Y(s)
Answer
Assuming that the order of six the plots is
(i=1) (i=2)
(i=3) (i=4)
(i=5) (i=6)
And that:
- From the first plot, the initial phase is 90º leading (a regular sine would start at 0º)
- In the fourth plot, phase lags 90º, so phase become 0º
- In the firth plot, it lags an additional 90º, so phase become 180º
Then, also considering the amplitudes, we could sketch a bode plot (like Andy aka suggested - took 5 minutes):
From this, it seems that structure is an derivator (DC gain = 0, positive magnitude slope and phase 90º leading) with two poles (same slope but descending, adds 180º of lag):
Y(s)=as(s+b)(s+c)U(s)
(It could actually be a double/triple/"n" integrator with four/six/"2n" poles, you can discover that by checking if the ascending or descending slope of the gain plot is 20 or 40 or 20*n db/decade).
You can then discover the coefficients by substituting $Y(s)$ and $U(s)$ in the preceding equation by the inputs and outputs (outputs will be the magnitude gain plus the phase lag), also substituting $s=jw$.
Hopefully then you will arrive at your answer.
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