The key thing to note here is that $E_\mathrm{cell} = E^\circ$ only when the ions involved are all in their standard states, i.e. concentrations of all ions is $\pu{1 M}$. But here we need to find $\Delta G^\circ$ for the reaction:
$$\ce{Zn(s) + Ag2O(s) + H2O(l) -> Zn^{+2}(aq) + 2Ag(s) + 2OH-(aq)},$$
whereas the redox reactions involved in the cell would be:
\begin{align}
\ce{Zn &-> Zn^{+2} + 2e-},\\
\ce{Ag+ + e- &-> Ag},
\end{align}
which requires $\ce{Ag+}$ to be present in its standard state. However, the first reaction given only tells us that $\ce{Ag2O}$ and $\ce{H2O}$ on the side of reactants are in their standard states. We hence don't know anything yet about the concentration of $\ce{Ag+}$, required in the cell.
Using the expression of $K_\mathrm{sp}$, $\ce{[Ag+][OH-]} = \pu{2E-8 M2}$:
From the products side of the first reaction, we know that $\ce{OH-}$ is in its standard state, so $\ce{[OH-] = 1 M}$, which gives $\ce{[Ag+]} = \pu{2E-8 M}$.
Now we use Nernst equation for this cell:
\begin{align}
\Delta G^\circ &= -nFE_\mathrm{cell}\\
\Delta G^\circ &= -nF\left(E^\circ_\mathrm{cell}
- \frac{0.059}{n}\log\frac{\ce{Zn^{+2}}}{\ce{Ag+}}\right)
\end{align}
$\ce{[Zn^{+2}]} = \pu{1 M}$ as it is in its standard state. Using $\ce{[Ag+]} = \pu{2E-8 M}$, we get:
$$ \Delta G^\circ = -2 \times 96.5 \times (1.56 - 0.059(8-\log2)) = \pu{-213 kJ/mol}$$
\pu{2\times 10^{-8} M^2}->\pu{2E-8 M2}$\pu{2E-8 M2}$ $\endgroup$