= Solution
Urn $i$ has $2^i$ balls, so a draw from it is white with probability $2^{-i}$. Given that the first ball is white, <Bayes' theorem> gives the <posterior probability>
$$
\mathbb P(I=i\mid W)=\frac{2^{-i}}{\sum_{j=1}^n2^{-j}}.
$$
After replacement, the second draw is <conditionally independent> of the first given the urn. Hence
$$
p(n)=\frac{\sum_{i=1}^n4^{-i}}{\sum_{i=1}^n2^{-i}}
=\frac{\frac13(1-4^{-n})}{1-2^{-n}}
=\frac13(1+2^{-n}).
$$
Consequently $\boxed{p(n)\to1/3}$.
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