I think Barro means in the footnote that Giovanni and Weil find the same equation, $U_t=\Phi C^{1-\gamma}$, but using the optimal path of $C_t$.
In Barro's paper, the approach is different given that the dynamics of $C_t$ is exogenous: $C_t=Y_t$ by assumption.
Barro uses the limit case when the length of a period gets close to 0. Maybe what may bother the reader is that the model is defined as discrete.
Rewrite the model
First, we can rewrite the model with a length of period $\delta$ and then use $\delta\to 0$.
The GDP dynamics write
$$\log(Y_{t+\delta})=\log(Y_t)+g\delta+u_{t+\delta}+v_ {t+\delta}$$
with $u_{t+\delta}\sim \mathcal{N}(0,\delta\sigma^2)$, and $v_{t+\delta}=0$ with probability $1-p\delta$ and $\log(1-b)$ with probability $p\delta$.
The utility satisfies
$$
U_t=\frac{1}{1-\gamma}\left\lbrace C_t^{1-\theta}+\frac{1}{1+\rho\delta}\left[(1-\gamma)E_tU_{t+\delta}\right]^\frac{1-\theta}{1-\gamma}\right\rbrace^\frac{1-\gamma}{1-\theta}.
$$
1) Find $\Phi$ as a function of $E_t\left[\left(\frac{C_{t+\delta}}{C_t}\right)^{1-\gamma}\right]$
From now suppose there is a $\Phi$ such that $U_t=\Phi C^{1-\gamma}$ (note that $\Phi$ depends on $\delta$ a priori).
Define $H(U)=[(1-\gamma)U]^\frac{1-\theta}{1-\gamma}$, the utility satisfies
\begin{align}
H(U_t)= C_t^{1-\theta}+\frac{1}{1+\rho\delta}H(E_tU_{t+\delta}).
\end{align}
We substitute $U_t$:
\begin{align}
H(\Phi)C_t^{1-\theta}= C_t^{1-\theta}+\frac{1}{1+\rho\delta}H(\Phi)\left(E_t\left[C_{t+\delta}^{1-\gamma}\right]\right)^\frac{1-\theta}{1-\gamma}.
\end{align}
Hence, we obtain for $C_t\neq 0$,
\begin{align}
\frac{1}{H(\Phi)}= 1-\frac{1}{1+\rho\delta}\left(E_t\left[\left(\frac{C_{t+\delta}}{C_t}\right)^{1-\gamma}\right]\right)^\frac{1-\theta}{1-\gamma}.
\end{align}
2) Find $E_t\left[\left(\frac{C_{t+\delta}}{C_t}\right)^{1-\gamma}\right]$ fromp the GDP dynamics
The trick is to find the expectation in the right-hand side from the GDP dynamics.
\begin{align}
\left(\frac{Y_{t+\delta}}{Y_t}\right)^{1-\gamma}= \exp\left((1-\gamma)g\delta\right).\exp\left((1-\gamma)u_{t+\delta}\right).\exp\left((1-\gamma)v_ {t+\delta}\right).
\end{align}
Taking the expectation and using the independence between $u_{t+1}$ and $v_{t+1}$, it follows
\begin{align}
E_t\left(\frac{Y_{t+\delta}}{Y_t}\right)^{1-\gamma}= \exp\left((1-\gamma)g\delta\right).E_t\exp\left((1-\gamma)u_{t+\delta}\right).E_t\exp\left((1-\gamma)v_ {t+\delta}\right).
\end{align}
The expectation of $\exp(X)$ where $X$ follows $\mathcal{N}(0,\sigma^2)$ is $\exp(\sigma^2/2)$. $\exp\left((1-\gamma)v_ {t+\delta}\right)$ is a random variable equal to $1$ with probability $1-p\delta$ and $(1-b)^{1-\gamma}$ with probability $p\delta$.
We substitute the expectation operator:
\begin{align}
E_t\left(\frac{Y_{t+\delta}}{Y_t}\right)^{1-\gamma}= \exp\left((1-\gamma)g\delta\right).\exp\left(\frac{(1-\gamma)^2\sigma^2\delta}{2}\right).\left(1-p\delta+pE[(1-b)^{1-\gamma}]\delta\right).
\end{align}
Finally, we use $C_t=Y_t$ to compute an equation for $\Phi$:
\begin{align}
\frac{1}{H(\Phi)}&= 1-\frac{1}{1+\rho\delta}\left\lbrace\exp\left((1-\theta)g\delta\right).\exp\left(\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\right)\right.\\
&\left. .\left(1-p\delta+pE[(1-b)^{1-\gamma}]\delta\right)^\frac{1-\theta}{1-\gamma}\right\rbrace.
\end{align}
3) Take the approximation $\delta\to 0$
The last step consists in taking a first-order approximation (I abusively keep the equal symbol):
\begin{align}
\frac{1}{H(\Phi)}&= 1-(1-\rho\delta). \left(1+(1-\theta)g\delta\right).\left(1+\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\right)\\
& .\left(1-\frac{1-\theta}{1-\gamma}p\delta+\frac{1-\theta}{1-\gamma}pE[(1-b)^{1-\gamma}]\delta\right).
\end{align}
Pursuing the first-order apprixmation (all the $\delta^i$ with $i>1$ can be neglected), we have
\begin{align}
\frac{1}{H(\Phi)}&= \rho\delta -(1-\theta)g\delta-\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\\
& +\frac{1-\theta}{1-\gamma}p\delta-\frac{1-\theta}{1-\gamma}pE[(1-b)^{1-\gamma}]\delta.
\end{align}
Substitute $g$ using $g^*=g+\frac{\sigma^2}{2}-pEb$,
\begin{align}
\frac{1}{H(\Phi)}&= \rho\delta -(1-\theta)g^*\delta+(1-\theta)\frac{\sigma^2}{2}\delta -(1-\theta)pEb\delta -\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\\
& +\frac{1-\theta}{1-\gamma}p\delta-\frac{1-\theta}{1-\gamma}pE[(1-b)^{1-\gamma}]\delta.
\end{align}
We take $\delta=1$ and invert function $H$ to find the solution in the footnote 7 of the paper. The right-hand side of this equation "simplifies" to the within braces in the formula.
