= Solution
At <pH> $3$, $[\mathrm{H^+}]=10^{-3}\,\mathrm M$, so the nitrate <concentration> is $0.001+2(0.1)=0.201\,\mathrm M$. With ideal <thermodynamic activities>,
$$
Q=\frac{(0.1)^3(0.1)^2}{(0.201)^2(10^{-3})^8}=2.47519\times10^{20}.
$$
Hence
$$
\boxed{E=1.719-\frac{8.3145(298)}{6(96485)}\ln(2.47519\times10^{20})=1.5180\,\mathrm V.}
$$
The activity of nitric oxide is $0.1$ because the supplied pressure is $0.1\,\mathrm{bar}$ and the gas standard is $1\,\mathrm{bar}$. This numerical result additionally approximates zinc and nitrate <thermodynamic activities> by their standard-concentration ratios; without their <activity coefficients>, a more precise nonideal value is not specified.
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