@vijju532
It specifically mentioned box A has atleast 2 letters. Its not right to consider "one box".
Let the 5 letters be named as L1,L2,L3,L4,L5.
Case 1) Put 2 letters in A. Ways of choosing 2 letters from 5 is C(5,2).
No. of ways of putting remaining 3 letters into B and C is $2^3$.
Let A has L1 and L2. So remaining is L3 to L5.
L3 can go to B or C. 2 ways
L4 can go to B or C. 2 ways.
L5 can go to B or C. 2 ways.
So total cases here is $C(5,2)x2^3 =10*8=80$
Case 2)
Put 3 letters in A. Ways of choosing 3 letters from 5 is C(5,3).
No. of ways of putting remaining 2 letters into B and C is $2^2$.
So total cases here is $C(5,3)x2^2 =10*4=40$
Case 3)
Put 4 letters in A. Ways of choosing 4 letters from 5 is C(5,4).
No. of ways of putting remaining 1 letters into B and C is $2^1$.
So total cases here is $C(5,4)x2^1 =5*2=10$
Case 4)
Put 5 letters in A. Ways of choosing 5 letters from 5 is C(5,5).
No. of ways of putting remaining 5 letters into B and C is $2^0$.
So total cases here is $C(5,5)x2^0 =1*1=1$
Sum them up: 80+40+10+1=131