Question pertaining to alkaline hydrolysis of ester.?

The reaction is second-order:  r = k₂[Ester][OH⁻]
As [Ester]ₒ = [OH⁻] and mole ratio Ester: OH⁻ = 1 : 1, [Ester] = [OH⁻] at all time t.
Hence, the rate equation can be written as:  r = k₂[Ester]²
The integrated form of the rate law:  1/[Ester] = kt + (1/[Ester]ₒ)

When t = 2 min, [Ester] = (1 - 20%)[Ester]ₒ = 0.8[Ester]ₒ:
1/(0.8[Ester]ₒ) = k(2) + (1/[Ester]ₒ)
1.25/[Ester]ₒ = 2k + (1/[Ester]ₒ)
2k = 0.25/[Ester]ₒ
k = 0.125/[Ester]ₒ …… [1]

When [Ester] = (1 - 80%)[Ester]ₒ = 0.2[Ester]:
1/(0.2[Ester]ₒ) = kt + (1/[Ester]ₒ)
5/[Ester]ₒ = kt + (1/[Ester]ₒ)
kt = 4/[Ester]ₒ …… [2]

[2]/[1]:
t = 4/0.125 min
Time taken to convert 80% of the ester, t = 32 min

The answer: c. 32 minutes