(i) The rate of change of the population is given by \(\frac{dN}{dt} = kN(1 - 0.01N)\). Given \(\frac{dN}{dt} = 0.32\) when \(N = 20\), we have:

\(0.32 = k \times 20 \times (1 - 0.01 \times 20)\)

\(0.32 = 20k \times 0.8\)

\(0.32 = 16k\)

\(k = 0.02\)

(ii) Separate variables and integrate:

\(\int \frac{1}{N(1 - 0.01N)} \, dN = \int 0.02 \, dt\)

Use partial fraction decomposition:

\(\frac{1}{N(1 - 0.01N)} = \frac{A}{N} + \frac{B}{1 - 0.01N}\)

Solving gives \(A = 1\) and \(B = 0.01\).

Integrate:

\(\int \left( \frac{1}{N} + \frac{0.01}{1 - 0.01N} \right) \, dN = \int 0.02 \, dt\)

\(\ln N - \ln(1 - 0.01N) = 0.02t + C\)

At \(t = 0, N = 20\):

\(\ln 20 - \ln(1 - 0.2) = C\)

\(C = \ln 25\)

Thus, \(\ln N - \ln(1 - 0.01N) = 0.02t + \ln 25\)

Rearrange to find \(t\):

\(t = 50 \ln \left( \frac{4N}{100 - N} \right)\)

(iii) When the population doubles, \(N = 40\):

\(t = 50 \ln \left( \frac{4 \times 40}{100 - 40} \right)\)

\(t = 50 \ln 4\)

\(t = 49.0\)