The magnetic field of an E.M. wave is given by $$\vec{B} = \left(\frac{\sqrt{3}}{2}\,\hat{i} + \frac{1}{2}\,\hat{j}\right) 30 \sin\left[\omega \left(t - \frac{z}{c}\right)\right] \; \text{(S.I. Units)}.$$ corresponding electric field in S.I. units is :

We have magnetic field $$\vec{B} = \Bigl(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\Bigr)30\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$$

For an electromagnetic wave propagating along +z, vectors $$\vec{E}$$, $$\vec{B}$$, $$\hat{k}$$ are mutually perpendicular and satisfy the right hand rule $$\vec{E}\times\vec{B}=+\hat{k}$$ (1), and their magnitudes satisfy $$|\vec{E}|=c\,|\vec{B}|$$ (2). Let $$\vec{E} = (E_x\hat{i}+E_y\hat{j})\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$$

The orthogonality condition $$\vec{E}\cdot\vec{B}=0$$ gives
$$\Bigl(\frac{\sqrt{3}}{2}E_x + \frac{1}{2}E_y\Bigr)30 = 0\,, $$
which simplifies to $$\frac{\sqrt{3}}{2}E_x + \frac{1}{2}E_y = 0\,$$ (3), so $$E_y = -\sqrt{3}\,E_x.$$

Using the magnitude relation (2) gives
$$E_x^2 + E_y^2 = (30c)^2$$
and substituting $$E_y = -\sqrt{3}E_x$$ yields
$$4E_x^2 = 900c^2\,, $$ so $$E_x^2 = 225c^2$$ and $$E_x = \pm15c,\quad E_y = \mp15\sqrt{3}\,c$$ (4).

The right hand rule (1) requires the z‐component of $$\vec{E}\times\vec{B}$$ to be positive, which gives
$$E_x\Bigl(\frac{30}{2}\Bigr) - E_y\Bigl(\frac{30\sqrt{3}}{2}\Bigr) > 0.$$
Substituting $$E_x = 15c,\;E_y = -15\sqrt{3}c$$ yields a positive value, so these signs are chosen.

Therefore the electric field is
$$\vec{E} = \Bigl(\tfrac{1}{2}\hat{i} - \tfrac{\sqrt{3}}{2}\hat{j}\Bigr)\,30c\;\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$$
This matches Option A.