Given${\mathrm{E}}_{\mathrm{Cr}+3/\mathrm{Cr}}^0=-0.72\mathrm{V};{\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^0=-0.42 V$
The potential for the cell, $\mathrm{Cr}\left|{\mathrm{Cr}}^{3+}\left(0.1\mathrm{M}\right)\parallel {\mathrm{Fe}}^{2+}\left(0.01\mathrm{M}\right)\right|\mathrm{Fe}$ is 

$\begin{array}{l}{\mathrm{E}}_{cell }^o={\mathrm{E}}_{cathode }^o-{\mathrm{E}}_{anode }^o\\ {\mathrm{E}}_{cell }^o=(-0.42)-(-0.72)\\ =+0.30\mathrm{V}\end{array}$
For the given galvanic cell the reaction is written as
$2\mathrm{Cr}+3{\mathrm{Fe}}_{(0.01\mathrm{M})}^{2+}\leftrightharpoons 2{\mathrm{Cr}}_{(0.1\mathrm{M})}^{3+}+3\mathrm{Fe}$
The reaction $\text{ Quitient (Qc) }=\frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{{\left[{\mathrm{Fe}}^{2+}\right]}^3}=\frac{[0.1{]}^2}{[0.01{]}^3}={10}^4$
The Nersnt equation is written as ${\mathrm{E}}_{\text{cell }={\mathrm{E}}_{\text{cell }}^o}-\frac{0.0591}{\mathrm{n}}$
$\log {\mathrm{Qc}}_{\mathrm{cell}}=0.30-\frac{0.0591}{6}\log {10}^4$ on solving ${\mathrm{E}}_{\text{cell }}=+0.261\mathrm{V}$