= Solution
The chain is a <birth-death chain>. Detailed balance between $0$ and $1$ gives
$$
\pi_0p=\pi_1q^{-1},
\qquad\text{so}\qquad
\pi_1=qp\,\pi_0.
$$
For $i\geq1$, detailed balance between $i$ and $i+1$ gives
$$
\pi_iq^{-(i+2)}=\pi_{i+1}q^{-(i+1)},
\qquad
\pi_{i+1}=\frac{\pi_i}{q}.
$$
Consequently
$$
\pi_i=qp\,\pi_0q^{-(i-1)},\qquad i\geq1.
$$
The expected occupation time of the positive even states during a return cycle to state $1$ is therefore
$$
\frac{\pi_2+\pi_4+\cdots}{\pi_1}
=\sum_{r=1}^{\infty}q^{-(2r-1)}
=\boxed{\frac{q}{q^2-1}}.
$$
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