The electric field!

We are taking two parts of the figure. Electric field

We take a elemental line charge of width dx .

So we have ${ dE }_{ 1 }=\frac { k\lambda }{ x } (sin{ 45 }^{ 0 }+sin{ 45 }^{ 0 })=\surd 2\frac { k\lambda }{ x }$

Also $\lambda =\sigma (dx)$

$\Rightarrow { dE }_{ 1 }=\sqrt { 2 } k\sigma \frac { dx }{ x }$

$\Rightarrow { E }_{ 1 }=(\sqrt { 2 } k\sigma )\int _{ b }^{ a }{ \frac { dx }{ x } } =\sqrt { 2 } k\sigma (ln(\frac { a }{ b } ))\hat { i }$

Similarly due to the other part ${ E }_{ 2 }=\sqrt { 2 } k\sigma (ln(\frac { a }{ b } ))\hat { j }$

${ E }_{ net }=\sqrt { 2 } k\sigma (ln(\frac { a }{ b } ))\hat { i } +\sqrt { 2 } k\sigma (ln(\frac { a }{ b } ))\hat { j }$

$\Rightarrow \left| { E }_{ net } \right| =2k\sigma (ln(\frac { a }{ b } ))=\frac { \sigma }{ 2{ \pi \varepsilon }_{ 0 } } (ln(\frac { a }{ b } ))$

Putting the values we have $\boxed { { E }_{ net }=802.08 }$