Electric Field & Field Lines question from JEE Main 2021 (27 Jul, Shift 2)

Answer: (b)

1. Take $O$ as origin, $E_0 = \dfrac{q}{4\pi\varepsilon_0 l^2}$. Side charges (distance $l$): $B(+q)$ and $G(2q)$ give $-E_0\hat{i}$ net; $C(2q)$ and $F(q)$ give $-E_0\hat{j}$ net.
2. Corner charges at distance $\sqrt{2}l$ give fields $\dfrac{E_0}{2}\cdot\dfrac{\text{charge}}{q}$ along the diagonals.
3. $A(-q)$, $D(+q)$, $H(-q)$, $E(2q)$ together: $\dfrac{E_0}{2\sqrt{2}}(\hat{i} + \hat{j})$
4. $\vec{E} = E_0\left(-1 + \dfrac{1}{2\sqrt{2}}\right)(\hat{i} + \hat{j})$, so $|\vec{E}| = E_0\left(\sqrt{2} - \dfrac{1}{2}\right) = \dfrac{q}{4\pi\varepsilon_0 (2l^2)}(2\sqrt{2} - 1)$
5. So the answer is (b).