digital low pass Butterworth filter
I'll try to give you some hints to get you started. First of all, you should know that the bilinear transform is given by

$$s=k\frac{z-1}{z+1}\tag{1}$$

If the analog prototype filter is normalized such that its cut-off frequency is $\Omega_c=1$, then the constant $k$ in (1) is given by

$$k=\frac{1}{\tan\left(\frac{\omega_0}{2}\right)}\tag{2}$$

where $\omega_0$ is the desired cut-off frequency of the discrete-time filter. Some more background on this is given in this answer.

You should also know that the poles of the normalized analog filter lie on a circle with radius $1$ centered at $s=0$ (of course the poles only lie on the left half of the circle). If the filter order is odd as in your example (why?), there must be a pole at $s=-1$. Now you can try to figure out how that pole is transformed to the $z$-plane, i.e.

$$\frac{1}{1+s}{\huge|}_{s=k\frac{z-1}{z+1}}=\ldots\tag{3}$$

From (3), knowing that that pole is transformed to a pole at $z=0.6$, you can determine the constant $k$, and from that, via Eq. (2), you can compute $\omega_0$. Now you know that the filter's attenuation is $3\text{ dB}$ at $\omega_0$, and it is $44\text{ dB}$ at $\omega=1.2$.

The last thing you should know is that with a Butterworth lowpass filter of order $N$, you get approximately $6N\text{ dB}$ attenuation per octave, at least in the frequency range considered here. Now figure out how many octaves there are between $\omega_0$ and $\omega=1.2$, figure out how many dB's difference in attenuation there must be between those two frequencies, and from this compute an estimate of the filter order $N$.