stochastic calculus - Using Black-Scholes to price a geometric average price call

Sorry if this is the wrong exchange for this question. It seems to be the most relevant, anyway.

I'm trying to learn and understand the Black-Scholes framework, with a focus on the stochastic differential equation approach (the exam I will be taking focuses on it). So I set out a challenge for myself. I'd like to price a special geometric average price call, where the average is taken on $S_0$ and $S_1$.

My intuition is that what I'm "really" trying to price is a European call, where the underlying is the geometric average of the stock price. I defined a process $G(t)$ by

\begin{equation*} G(t) = \left(S_0 S_t\right)^{\frac{1}{2}}. \end{equation*}

The intention is to apply Ito's lemma, so I took derivatives: \begin{align*} G_t &= 0 & G_S &= \frac{1}{2}S_0S_t^{-\frac{1}{2}} & G_{SS} = -\frac{1}{4}S_0^{-\frac{3}{2}}. \end{align*}

After applying Ito's lemma, I end up with the stochastic differential equation \begin{equation*} \frac{\mathrm{d}G(t)}{G(t)} = \frac{1}{2}\left[\left(\alpha - \delta - \frac{1}{4}\sigma^2\right) \mathrm{d}t + \sigma \mathrm{d} Z_t \right], \end{equation*} where $\alpha$ is the stock's expected rate of return.

So I see that $G(t)$ is a geometric Brownian motion. But this is where I become deeply confused, since it is a derivative of the stock $S_t$. So when I do risk-neutral pricing, do I have to assume that $S$ earns the risk-free rate (which amounts to setting $\alpha = r$, in the stochastic differential equation above), or do I assume that G earns the risk-free rate? Or something else?

My intuition is telling me that once I figure out which rates to use and where, I can just use the Black-Scholes formula for a call to get this claim price done. Am I on the right track?