Due to wire along z-axis,

$\overrightarrow {E_3}=\large\frac{\lambda}{2 \pi \in _0 (a \sqrt 2)} $$\quad ( \cos 45^{\circ} \hat i + \cos 45^{\circ} \hat j)$

=> $\overrightarrow {E_3} =\large\frac{\lambda}{4 \pi \in _0 a} $$(\hat i +\hat j)$

Similarly,

$E_1=\large\frac{\lambda}{4 \pi \in_0 a}$$(\hat j+\hat k)$

=> $\overrightarrow {E_2} =\large\frac{\lambda}{4 \pi \in _0 a} $$(\hat i +\hat k)$

=> $\overrightarrow {E}_{resultant } =\overrightarrow {E_1}+\overrightarrow {E_2}+\overrightarrow {E_3}$

$\large\frac{\lambda} {2 \sqrt 2 \pi \in _0 a} (\hat i +\hat j + \hat k)$

Hence B is the correct answer.