= Solution
For monovalent mobile ions, the linearized <Debye–Hückel screening length> is
$$
\lambda^2=\frac{\epsilon_0\epsilon_r k_BT}{e^2\sum_i n_i z_i^2}.
$$
Here $n_i$ is a <number density> in $\mathrm{m^{-3}}$, not a <molarity>. Use the question's idealized neutral-water convention, $c_+=c_-=10^{-7}\,\mathrm{mol\,L^{-1}}$, and $T=293.15\,\mathrm K$. Both ion species contribute:
$$
\lambda=\sqrt{\frac{80\epsilon_0k_BT}{2e^2N_A\,10^3\times10^{-7}}},
\qquad \boxed{\lambda_{\rm water}\simeq9.63\times10^{-7}\,\mathrm m\simeq0.96\,\mu\mathrm m.}
$$
For added salt concentration $c_s$ in $\mathrm{mol\,L^{-1}}$, the <ionic strength> in this model is $I=c_s+10^{-7}\,\mathrm{mol\,L^{-1}}$. With the numerical <molarity> $I_M=I/(1\,\mathrm{mol\,L^{-1}})$,
$$
\lambda=\frac{0.3045\,\mathrm{nm}}{\sqrt{I_M}},\qquad
\boxed{c_s\simeq0.0927\,\mathrm{mol\,L^{-1}}\simeq0.093\,\mathrm M\quad(\lambda=1\,\mathrm{nm}).}
$$
The water contribution is negligible at that salt level. The factor $10^3$ converts molarity into $\mathrm{mol\,m^{-3}}$; omitting it or counting only one ion species gives a wrong screening <length>. Exact water autoionization and activities vary with <temperature>, so the first estimate uses the simplified neutral $10^{-7}\,\mathrm M$ concentrations implied by the question. At the higher salt level, ion-size effects, nonideal activities and changes in <permittivity> limit the numerical accuracy of this continuum estimate.
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