The half-life of the element is
You can actually solve this problem without using the half-life equation.
When you're dealing with a sample of a radioactive element, you know that its initial mass will be halved with every half-life that passes.
In other words, you are left with half of what you started with after one half-life, you are left with a quarter of what you starter with after two half-lives, you are left with one eighth of what you started after three half-lives, and so on.
You can actually write this as
#A = A_0/2^(n)#, where
In your case, you know that you're left with
#underbrace(1/8 * color(red)(cancel(color(black)(A_0))))_(color(blue)(1/8"th of the original sample")) = color(red)(cancel(color(black)(A_0))) * 1/2^n#
This is equivalent to
#1/8 = 1/2^n#
#1/2^3 = 1/2^n implies n = 3#
So, three half-lives have passed in 90 seconds, which can only mean that
#3 * t_"1/2" = "90 s" implies t_"1/2" = "90 s"/3 = color(green)("30 s")#