Verified answer

approximately 1 month is the key

one month is approx 4 * 8 days

I am sure this is how they want you to do this considering your other problems

I doubt if you know logs

2^4=16

2 E9/16 = 20 E9/16= 1.25 E9 atoms

N(t)=N0 * e^(-t/T)

where N(t) is the number of atoms at time t, N0 initial number, T is the mean life

T=t_half_life / ln(2)

N(30) = 2*10^10 * e^(-30*ln(2)/8) =~ 2*10^10 * e^(-2.6) =~ 2*10^10 * 0.074 =~1.48*10^9 [atoms]

Alternatively you can do a crude estimation by half-time periods.

30 is between 8*3=24 and 8*4=32 but closer to 32, so 4 half-time periods

After 4 half-time periods you'll have 1/2*1/2*1/2*1/2 = 1/16 of the original quantity

1/16 * 2*10^10=1.25*10^9 - but keep in mind that your actual number should be higher.

However it will be lower than after 3 half-periods -> 1/8 * 2*10^10 = 2.5*10^9