Galvanic cell involving a hydrogen electrode and copper electrode in sodium hydroxide
In this case, "saturated" means that there is enough dissolved copper (II) hydroxide such that no more will dissolve, i.e. the equilibrium represented by the $K_\mathrm{sp}$, $\ce{Cu(OH)2_{\ (s)}&lt;=&gt;Cu^2+_{(aq)} + 2OH-_{(aq)}}$ holds true. So first, let's figure out the concentration of copper (II) hydroxide that will dissolve in 0.1 M NaOH (making the assumption, as mentioned in the comments, that $[\ce{OH^-}] = \mathrm{C_\ce{NaOH}})$:


The $K_\mathrm{sp}$ is really really small, so we can actually ignore
its contribution to the $[\ce{OH^-}]$, but for fun, you can do the
whole ICE table thing and end up with:
$$\begin{align}
     [\ce{OH^-}] &amp;= .1 + 2x \\
     [\ce{Cu^2+}] &amp;= x \\
     K_\mathrm{sp} &amp;= [\ce{OH^-}]^2[\ce{Cu^2+}] \\
     1.6\times 10^{-19}  &amp;= (.1 + 2x)^2(x) \\
     x = [\ce{Cu^2+}] &amp;= 1.6 \times 10^{-17}\ \mathrm{M}
     \end{align}
   $$


Next, using the Nernst equation, we can calculate the copper half-cell reduction potential (Assuming T = 298.15 K):


$$\begin{align}
    E &amp;= E^\circ - \frac{RT}{zF}\ln\frac{1}{[\ce{Cu^2+}]} \\
      &amp;= 0.34\ \mathrm{V} - \frac{RT}{2F}\ln\frac{1}{1.6 \times 10^{-17}} \\
      &amp;= -0.16\ \mathrm{V}
    \end{align}
    $$


Finally, the whole cell potential:


First, negate the calculated reduction potential to get the oxidation potential:
$$\begin{align}
    E_\mathrm{ox} &amp;= -E_\mathrm{red} \\
      &amp;= 0.16\ \mathrm{V}
    \end{align}
    $$
Then, because the SHE's reduction potential is defined as 0 V, the whole cell potential is the same as the oxidation half-cell potential:

$$\begin{align}
    E_\mathrm{ox} &amp;= E_{(\ce{H})\mathrm{red}} + E_{(\ce{Cu})\mathrm{ox}} \\
                  &amp;= 0\ \mathrm{V} + 0.16\ \mathrm{V} \\
                  &amp;= 0.16\ \mathrm{V}
    \end{align}
    $$