detailed solution

Correct option is A

$\begin{array}{l}\mathrm{Resistance} \mathrm{per} \mathrm{unit} \mathrm{length} = \frac{\mathrm{R}}{\mathrm{l}}=0.01\mathrm{\Omega }{\text{ m}}^{-1},\\ \mathrm{r}=1.0\mathrm{cm}\text{ Total turns }=400,\mathrm{\ell }=20\mathrm{cm},\\ \mathrm{B}=1\times {10}^{-2}\mathrm{T},\mathrm{n}=\frac{400}{20\times {10}^{-2}}\text{ turns }/\mathrm{m}\\ \mathrm{i}=\frac{\mathrm{E}}{{\mathrm{R}}_0}=\frac{\mathrm{E}}{\frac{\mathrm{R}}{\mathrm{l}}\times (2\mathrm{\pi r}\times 400)}=\frac{\mathrm{E}}{0.01\times 2\times \mathrm{\pi }\times 0.01\times 400}\\ \mathrm{B}={\mathrm{\mu }}_0\mathrm{ni}\\ \Rightarrow {10}^{-2}=4\mathrm{\pi }\times {10}^{-7}\times \frac{400}{20\times {10}^{-2}}\times \frac{\mathrm{E}}{400\times 2\mathrm{\pi }\times 0.01\times {10}^{-2}}\end{array}$
$\Rightarrow \mathrm{E}=\frac{{10}^{-2}\times 20\times {10}^{-2}\times 400\times 2\mathrm{\pi }\times {10}^{-2}\times 0.01}{4\mathrm{\pi }\times {10}^{-7}\times 400}=1\mathrm{V}$

detailed solution

Correct answer is 1

$\begin{array}{l}\mathrm{Resistance} \mathrm{per} \mathrm{unit} \mathrm{length} = \frac{\mathrm{R}}{\mathrm{l}}=0.01\mathrm{\Omega }{\text{ m}}^{-1},\\ \mathrm{r}=1.0\mathrm{cm}\text{ Total turns }=400,\mathrm{\ell }=20\mathrm{cm},\\ \mathrm{B}=1\times {10}^{-2}\mathrm{T},\mathrm{n}=\frac{400}{20\times {10}^{-2}}\text{ turns }/\mathrm{m}\\ \mathrm{i}=\frac{\mathrm{E}}{{\mathrm{R}}_0}=\frac{\mathrm{E}}{\frac{\mathrm{R}}{\mathrm{l}}\times (2\mathrm{\pi r}\times 400)}=\frac{\mathrm{E}}{0.01\times 2\times \mathrm{\pi }\times 0.01\times 400}\\ \mathrm{B}={\mathrm{\mu }}_0\mathrm{ni}\\ \Rightarrow {10}^{-2}=4\mathrm{\pi }\times {10}^{-7}\times \frac{400}{20\times {10}^{-2}}\times \frac{\mathrm{E}}{400\times 2\mathrm{\pi }\times 0.01\times {10}^{-2}}\end{array}$
$\Rightarrow \mathrm{E}=\frac{{10}^{-2}\times 20\times {10}^{-2}\times 400\times 2\mathrm{\pi }\times {10}^{-2}\times 0.01}{4\mathrm{\pi }\times {10}^{-7}\times 400}=1\mathrm{V}$