= Solution
Fix the horizon $T$ and define the <half-volatility measure for a square-root stock claim> by
$$
Z_s=\exp\left(\frac12\int_0^s\sigma_u\,dW_u-\frac18\int_0^s\sigma_u^2du\right),\qquad
\frac{dQ}{dP}\bigg|_{\mathcal F_T}=Z_T.
$$
The <Novikov condition> makes $Z$ a true <martingale> with $\mathbb EZ_T=1$, and $Z_T>0$ makes $Q$ equivalent to $P$. The <Girsanov theorem> identifies $W_s^Q=W_s-\frac12\int_0^s\sigma_u du$ as a <Brownian motion> under $Q$.
The exact factorization from the previous part is
$$
\sqrt{S_T}=\sqrt{S_t}\,\frac{Z_T}{Z_t}e^{-I_{t,T}/8}.
$$
Using the <Bayes formula for conditional expectation> therefore gives
$$
\boxed{C(t,T)=\sqrt{S_t}\,\mathbb E_Q[e^{-I_{t,T}/8}\mid\mathcal F_t].}
$$
The random integrated variance need not have any independence property under $Q$. Its pathwise bounds still imply $e^{-b^2\tau/8}\leq\mathbb E_Q[e^{-I_{t,T}/8}\mid\mathcal F_t]\leq e^{-a^2\tau/8}$. Inverting the same price function proves \b[$a\leq\Sigma(t,T)\leq b$ even for volatility adapted to the Brownian motion], on the positive-price event. The zero-price convention is as above. The auxiliary measure here is not asserted to be an <equivalent martingale measure> for the stock.
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