There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. Determine the number of horizontal layers in the pile?

There is 1 ball on top, 3 balls in second layer, 6 balls in third layer,10 balls in fourth layer .......... and so on 
Total balls=8436
Sum of balls across all layers = S = 1+3+6+10+.........
S = 1+3+6+10+.......+$$t_n$$ 
S =     1+3+6+........+$$t_{n-1}\ +\ t_n$$
Subtracting both equations,
0 = 1+2+3+4+......... - $$t_n$$
$$t_n$$ = 1+2+3+4+.......... = $$\ \frac{\ n\left(n+1\right)}{2}$$
$$S_n=\Sigma\ t_n=\Sigma\ \ \frac{\ n\left(n+1\right)}{2}=\ \frac{\ 1}{2}\Sigma\ \left(n^2+n\right)$$
$$S_n=\ \frac{\ 1}{2}\left(\ \frac{\ n\left(n+1\right)\left(2n+1\right)}{6}+\ \frac{\ n\left(n+1\right)}{2}\right)=\ \frac{\ n\left(n+1\right)\left(n+2\right)}{6}=8436$$
n(n+1)(n+2)=50616
On solving, we get n=36.
Option B is the answer.