Can you please explain.

IMO B

Let # of 23 cents pencils be x &
# of 21 cents pencils be y

a) x+y = 6
No idea whats the value of x and y , so Insufficient.
b) 23x+21y = 130

only if x= 2 , y is an integer. i.e 23*2 +21 y =130 => y = (130-46)/21 = 84/21 = 4
Sufficient.

To understand this, get the maximum multiple of 23 closest to 130.
23*5 = 115
and minimum could be 0
so , between 0 and 5, only at x=2 , equation 23x+21y = 130 satisfies Y to be an integer.

So, B

x=23 cents pencil
y=21 cents pencil
to find x.
A) x+y = 6. Insufficient as y is not known
B) 23x+21y = 130
(as x,y are numbers, both have to be integers)
x = (130-21y)/23
if y=1, x= 109/23 (not an integer) thus wrong
if y=2, x=88/23 wrong
if y=3, x=67/23 wrong
if y=4, x=46/23 = 2 (integer can be true)
if y=5, x=25/23 wrong
if y=6, x=4/23 wrong
if y=7, x=-17/23 wrong.(x can't be -ve)
As we increase y, x will become more and more negative.
Thus only valid answer is when y=4, x=2
Thus no. of 23 cents pencil =2 and B alone is sufficient.