If $E°$ of calomel electrode $\left({\mathrm{Cl}}^{-}\left|{\mathrm{Hg}}_2{\mathrm{Cl}}_2\right|\mathrm{Pt}\right)$ is $0.27\mathrm{V}$ its potential when $\mathrm{KCl}=0.1\mathrm{M}$ would be 

The reaction is $\frac{1}{2}{\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(\mathrm{s}\right)+{\mathrm{e}}^{-}\to \mathrm{Hg}\left(\mathrm{l}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$ 

Hence, $E=E^{\circ }-\frac{RT}{E}\ln \left(\left[{\mathrm{Cl}}^{-}\right]/c^{\circ }\right)=0.27\mathrm{V}-\left(0.059\mathrm{V}\right)\log \left(0.1\right)=0.2759\mathrm{V}$