current I	B (mT)
0.5
1.0
1.5
2.0
length of coil
number of turns N
n = N/L
Note the rounding to the 0.001 place in the I column; most of your recorded uncertainties were rounded to that place.  For example you could have recorded 0.1320 +/1 (plus or minus) 0.0011453 for the current.
Recall the graph paper  has a least count of 0.002 mT. Thus, the above numbers can easily be plotted. If the last digit is even, the point lies exactly on a small division. If the last digit is odd, the point lies at the midpoint between two divisions.
Thus mbest =
mmax =
mmin =
Compute (mmax - mmin ) =
Compute LCy/(delta I) = (0.002)/(1.500) = 0.0013333 mT.
Choose delta m as the largest of the previous two:
Now the slope should be  equal to uo*n, where uo is the magnetic permeability  of free space  and n is the number of turns per meter of the slinky.  Note n = N/L, where N is the number of loops and L is the length.  T
Finally compare your results with the actual, accepted  measured  uo*nacc   .   Compute the percent difference which is the absolute value of the difference divided by the average of the two.
Also check if the accepted value falls within the range in a way defined  by this inequality:  | uo*nbest - uo*nacc | < delta m + (uo*N/L2)*delta L, where delta L = 0.003 m. Watch your sig. figs. !