Derive and calibrate a logistic population model

For the specified initial state, $0<I<3000$. Since

$\frac1{I(3000-I)}=\frac1{3000}\left(\frac1I+\frac1{3000-I}\right),$

integration gives $\ln|I/(3000-I)|=3000kt+C$, hence $I/(3000-I)=Ae^{3000kt}$. Solving, $I=3000Ae^{3000kt}/(1+Ae^{3000kt})$.

$30=3000A/(1+A)$ gives $A=30/2970=1/99$. Therefore $I=3000/(1+99e^{-3000kt})$, as required.

$240=3000/(1+99e^{-21000k})$ gives $e^{-21000k}=23/198$ and $k=-\ln(23/198)/21000$. For $I=2700$, $e^{-3000kt}=1/891$. Thus $t=7\ln(1/891)/\ln(23/198)\approx22.1$ days from the initial observation, rather than an additional $22.1$ days after day $7$.

Any one source assumption suffices: the total population remains $3000$; infected people remain infected (no recovery during the model); or individuals have equal opportunity of infection under homogeneous mixing. These state the idealised model assumptions rather than conclusions about a real disease.