here 2 conditions are given
1) $(ab)^{2}$= ccb
2) ccb>300
so clearly we "ab " will be greater than 17 bcoz (17)^2 = 289
and here last digit in ab , which b and last digit in ccb which is b are same ..
(18)^2= 324 here 8 not equal to 2
only unit digit 0, 1 , 5 , 6 give same unit digit when we square the number .
so $(20)^{2}$ = 400 but here c =4 , c=0 , b= 0 so c not equal to c .
$(21)^{2}$ = 441 , here b = 1 , and c= 4 which follow all conditions means 1 should be the value of b ..
so ans should be A)