first of all bn = 2^n/n! since (-2)^n = (-1)^n 2^n
you want 2^(n+1)/[(n+1)!] < .01
There is no analytic way of solving this--simply make a table of values
using your calculator and look for the smallest n
for which 2^(n+1)/[(n+1)!] < .01
Or if you prefer crank it aout by hand try n= 2 if that doesn't work try n = 3 etc.