For the solution one $\mathrm{pH}=3$ :-

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \left[{\mathrm{H}}^{+}\right]= {10}^{-\mathrm{pH}}={10}^{-3} \end{gathered}$$

For the second solution $\mathrm{pH}=6$ :-

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \left[{\mathrm{H}}^{+}\right]= {10}^{-\mathrm{pH}}={10}^{-6} \end{gathered}$$

According to the nernest equation :-

$$\begin{gathered} {{\mathrm{E}}^{}}_{\mathrm{cell}}{={\mathrm{E}}^{°}}_{\mathrm{Cell}}-\frac{0.0591}{\mathrm{n}}\log \frac{\left[\mathrm{Oxidation}\right]}{\left[\mathrm{Reduction}\right]} \\ {{\mathrm{E}}^{}}_{\mathrm{cell}}{={\mathrm{E}}^{°}}_{\mathrm{cell}}-\frac{0.0591}{1}\log \frac{{10}^{-3}}{{10}^{-6}} \\ {{\mathrm{E}}^{}}_{\mathrm{cell}}=0+0.059\times 3=0.177 \end{gathered}$$

Hence, If hydrogen electrons dipped in two solutions of $\mathrm{pH}=3$ and $\mathrm{pH}=6$ are connected by a salt bridge, the emf of the resulting cell is $0.177 \mathrm{V}$.