OK so you have a shaft of length 500 mm and diameter 50 mm. You need to reduce the diameter by 6 mm. For your material and setup, your cutting speed (that is, how fast the tool can travel over the shaft) is $\dot{c} = 16 m/min$, and your maximum depth of cut is 2 mm. Longitudinal feed is given as 0.2 mm/rev, transverse feed 0.1 mm/rev.
I am going to assume that machine set time, measurement time, tool change time, etc are all zero. I will also assume that there is no facing cut to be made, and that the original metal you would be using is greater than 500 mm, since you are parting it off. I am also going to assume that your machine can be set to ANY rpm.
Since your machine will reduce the radius, this means that your total longitudinal cuts should come out to $\Delta = 3mm$. This would be done most efficiently with two cuts, of $\delta_1 = 2mm$ and $\delta_2 = 1 mm$. You would then have a parting cut.
Pass 1:
Longitudinal pass cutting $\delta_1 = 2mm$, across length $l=500mm$, original diameter $d_0 = 50 mm$. First, find the spindle RPM:
$$\nu_1 = \frac{\dot{c}}{\pi d_0} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(50 mm)}=102 rpm$$
Your travel speed is your longitudinal feed $f_z = 0.2 mm/rev)$ times your spindle speed:
$$ \dot{z}_1=f_z \nu_1 = 20.4 mm/min$$
From there, you can get your tie taken:
$$ l = \dot{z}_1t_1 \implies t_1=\frac{l}{\dot{z}_1}=\frac{500 mm}{20.4 mm/min} = 24.51 min$$
Pass 2: Your second pass should be identical to the first, except that the depth of cut is 1 mm. Normally, reducing your depth of cut means you can adjust your cutting speed by a factor $F_d$, but we will disregard that here since you don't have enough information to find it.
Since your new diameter is $d_1=46mm$, we will recalculate the spindle RPM:
$$\nu_2 = \frac{\dot{c}}{\pi d_1} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(46 mm)}=111 rpm$$
$$\dot{z}_2=f_z\nu_2 = 22.2mm/min$$
$$t_2=\frac{l}{\dot{z}_2}=\frac{500 mm}{22.2 mm/min} = 22.52 min$$
Pass 3: This is the parting off. New diameter is $d_2=44 mm$, and transverse travel feed is $f_x=0.1 mm/rev$. Lets get our new spindle speed and travel speed:
$$\nu_3 = \frac{\dot{c}}{\pi d_2} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(44 mm)}=116 rpm$$
$$\dot{x}_3=f_x\nu_3=(0.1 mm/rev)(116 rpm)=11.6 mm/min$$
$$r = \frac{d}{2} = \dot{x}_3t_3 \implies t_3=\frac{d}{2 \dot{x}_3}=\frac{44 mm}{2(11.6 mm/min)}=1.90 min$$
Total time: Your total time is the sum of these three passes, since we are ignoring everything else:
$$t=t_1+t_2+t_3 = 24.51 min + 22.52 min + 1.90 min = 48.93 min$$
Based on the assumptions made to solve the problem (and, barring a math error) this should be the machining time.
