For copper half cell, the graph between reduction potential (Y-axis) and log $\left[{\mathrm{Cu}}^{2+}\right]$ is a straight line with the Y-intercept + 0.34V. Reduction potential of copper electrode with 0.01 M ${\mathrm{CuSO}}_4$ solution is

Given here,
${\mathrm{E}}_{\mathrm{red}}^0=0.34\mathrm{v} \mathrm{and} \left[{\mathrm{Cu}}^{2+}\right]=0.01\mathrm{M}$
We have to find out the reduction potential of electrode i.e. ${\mathrm{E}}_{\mathrm{red}}=?$
For that we have to use Nernst equation
${\mathrm{E}}_{\mathrm{red}}={\mathrm{E}}_{\mathrm{red}}^0-\frac{0.0591}{\mathrm{n}}\log \frac{1}{\left[{\mathrm{Cu}}^{2+}\right]}$

By putting values in equation we get,

${\mathrm{E}}_{\mathrm{red}}=0.34-\frac{0.0591}{2}\log \frac{1}{0.01}$

${\mathrm{E}}_{\mathrm{red}}=0.34-0.0295 \log {10}^2$

${\mathrm{E}}_{\mathrm{red}}=0.34-0.0295\times 2$

By solving this we get

Reduction Emf of electrode is

${\mathrm{E}}_{\mathrm{red}}=0.399\simeq 0.40\mathrm{V}$

Hence option A is correct.