Blank data sheet here
current I B (mT)
0.5 0.064
1.0 0.129
1.5  0.201
2.0 0.264
length of coil 0.750 (m) +/-0.003 (m)
number of turns N  82
n = N/L 109.33
Note the rounding to the 0.001 place in the B 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 might be calculated from numbers used like (0.265 - 0.065)/(2.000 - 0.500) = 0.200/1.500 =

0.133322 mT
 mmax might be calculated from numbers used like (0.269 - 0.061)/(2.000 - 0.500) = 0.208/1.500 =

0.138666 mT.
Thus mmin might be calculated from numbers used like (0.260 - 0.069)/(2.000 - 0.500) = 0.191/1.500 =

0.127333 mT.
Compute (mmax - mmin ) = (0.138666 - 0.133333)/2 = 0.0026665 mT. 
Compute LCy/(delta I) = (0.002)/(1.500) = 0.0013333 mT. 
Choose delta m as the largest of the previous two: delta m = 0.0026665 mT
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.  Thus uo*n = mbest = 0.133322 mT.
Finally compare your results with the actual, accepted  measured  uo*nacc   = 
(1.256637x10-6)*(109.33) = 0.1373923078 tm. 
Compute the percent difference which is the absolute value of the difference divided  by the average of the two  and multiplied by 100 to achieve a percentage.
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. !