d_hkl                        sin²Ï´=λ²/4d_hkl²Â Â Â                 (sin²Ï´/0.04543)=h² + k² + l²Â Â Â Â Â Â Â Â Â Â 

0.2088                  0.1363                                 3

0.1808                  0.1818                                 4

0.1278                  0.3639                                 8

0.1090                  0.5003                                 11

0.1044                  0.5454                                 12

0.09038               0.7277                                 16

0.08293               0.8643                                 19

0.08083               0.9098                                 20

These correspond to hkl values of (111),(200),(220),(311) so on..

The reflection maxima thus observed from planes for which hkl are either all odd or all even. So the unit cell belongs to the FCC arrangement.

b.

The larger Bragg angles can be measured more accurately than the smaller one, so more accurate value for sin²Ï´ is 0.9098.

For this the corresponding d_hkl value is 0.08083

So, d_hkl² = λ²/4 sin²Ï´ = 4a²/h² + k² + l²

or, (0.08083² = 4a²/20

or, a = 0.1807 nm = 180.7 pm

c. Table is not given in the question.