= Solution
Within month two, each individual contributes only the time spent at risk between times one and two. There are $104-6-1=97$ complete one-month contributions. The six event contributions are $0.08,0.16,0.22,0.50,0.63,0.72$, and the censored contribution is $0.69$. Consequently
$$
T_2=97+(0.08+0.16+0.22+0.50+0.63+0.72)+0.69
=100\text{ person-months}.
$$
The month-specific <likelihood> factor in a <piecewise-exponential survival model> is $\theta_2^6e^{-\theta_2T_2}$, giving
$$
\boxed{\widehat\theta_2=\frac6{100}=0.06\text{ per month}.}
$$
The eight individuals no longer at risk at time one contribute no month-two exposure. Using either all 112 individuals or all 104 as full-month observations would give the wrong denominator.
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