deriving the transfer function given bode plot

Your assumed transfer function is wrong. A Bode plot MUST show the LF(low frequency) and HF(high frequency) asymptotes, otherwise it's not giving the full picture. Hence, we must assume the Bode plot presented contains all the information - there are no surprises above or below the frequency range shown.

In this case the LF asymptote is a slope of -20 dB/dec.

There are two break frequencies: one pole at 100 rad/sec, and a double pole at 1000 rad/sec. There is not a break frequency at 10 rad/sec.

The magnitude of a transfer function in dB is $$Magnitude=20log_{10}|H(jw)|$$ where H(jw) is the transfer function. Seeing the slopes in the graph shown above, there are poles at 100 and two poles at 1000Hz frequencies. So the transfer function would be $$H(jw)=\frac{k}{(jw+100)(jw+1000)^2}$$ Observe that there are two poles at 1000Hz.

Now at dc frequency/near dc (0.1 rad/s), the gain is 20dB. Gain at 0.1 rad/s is similar to dc frequency considering the pole magnitudes.

Therefore, $$20dB=20log_{10}|H(jw)|$$ $$log_{10}k-log_{10}[100*1000^2]=1$$ Solving we get $$k=10^{9}$$