Calculating the concentration of copper(II) from a non-standard cell potential
You are on the right track with that equation. There is a problem with the question, though: you do not know which direction the reaction is proceeding in. Therefore, you get two possible answers.

One of the two possible ways to write the reaction that's occurring is

$$\ce{Cu^2+(aq) + H2(g) -&gt; Cu(s) + 2H+ (aq)}$$

For this reaction, the value of $E_\mathrm{cell}^\circ$ is given by $E^\circ(\ce{Cu^2+}/\ce{Cu}) - E^\circ(\ce{H+}/\ce{H2}) = \pu{+0.34 V}$. This seems to be the more logical direction for the reaction to proceed in, as we have a positive value of $E_\mathrm{cell}^\circ$. Consequently, to achieve the smaller potential difference of $\pu{0.195 V}$, all we need to do is to reduce the concentration of $\ce{Cu^2+}$ from the standard of $\pu{1 mol dm-3}$, so that the forward reaction becomes slightly less favourable. The corresponding value of $n$ is $2$. The reaction quotient can be approximated by

$$Q = \frac{([\ce{H+}]/c^\circ)^2}{([\ce{Cu^2+}]/c^\circ) \cdot (p(\ce{H2})/p^\circ)} = \frac{c^\circ}{[\ce{Cu^2+}]}$$

since the $\ce{H+}/\ce{H2}$ electrode is operating under standard conditions (and where $c^\circ = \pu{1 mol dm-3}$). The temperature is not specified, but we could assume it to be $\pu{298 K}$.

Therefore:

$$\begin{align}
E_\mathrm{cell} &amp;= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &amp;= \pu{0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{c^\circ}{[\ce{Cu^2+}]}\right) \\
[\ce{Cu^2+}] &amp;= \pu{1.246 \times 10^-5 mol dm-3}
\end{align}$$



If the reaction is actually running in the reverse direction, then you have the reaction

$$\ce{Cu(s) + 2H+(aq) -&gt; Cu^2+(aq) + H2(g)}$$

with $E^\circ_\mathrm{cell} = \pu{-0.34 V}$. Going through exactly the same steps as above (but taking care to invert the expression for $Q$, since the reactants and products are swapped), you get to

$$\begin{align}
E_\mathrm{cell} &amp;= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &amp;= \pu{-0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{[\ce{Cu^2+}]}{c^\circ}\right) \\
[\ce{Cu^2+}] &amp;= \pu{8.019 \times 10^-19 mol dm-3}
\end{align}$$

What's happening here? It turns out that as you reduce the concentration of $\ce{Cu^2+}$ further, $E_\mathrm{cell}$ drops more and more, until it actually becomes thermodynamically favourable for the reverse reaction to occur. This concentration isn't particularly sensible, but you can certainly achieve it by serial dilution.



How do we resolve this dilemma? The most straightforward way is to find out which electrode is the anode and which is the cathode, which you can do with the voltmeter. For voltmeters to display a positive reading, as is suggested by the question, you need to have electrons flowing from the black wire to the red wire through the voltmeter. This means that the black wire must be connected to the electrode which provides it with electrons, i.e. the anode, since the anode is the site of oxidation (the chemical species gives up its electrons to the electrode and is itself oxidised in the process).

So, if your voltmeter displays a positive reading and the black wire is connected to the standard hydrogen electrode, it means that $\ce{H2}$ is being oxidised to $\ce{H+}$. That means that we are in the first situation above, where the forward reaction $\ce{Cu^2+(aq) + H2(g) -&gt; Cu(s) + 2H+ (aq)}$ is occurring.