If the magnetic field in a plane electromagnetic wave is given by $$\vec{B} = 3 \times 10^{-8} \sin(1.6 \times 10^3 x + 48 \times 10^{10} t)\hat{j}\,T$$, then what will be the expression for electric field?

We are given the magnetic field of a plane electromagnetic wave as

$$\vec B \;=\; 3 \times 10^{-8}\,\sin\!\bigl(1.6\times10^{3}\,x\;+\;48\times10^{10}\,t\bigr)\,\hat j\;\text T.$$

For any plane electromagnetic (EM) wave in free space we always have three mutually perpendicular vectors: the electric field $$\vec E,$$ the magnetic field $$\vec B,$$ and the direction of wave propagation $$\vec k.$$ The following two vector relations hold:

$$\vec E \perp \vec B,\qquad \vec E \times \vec B \;\text{is along the direction of propagation}.$$

First we determine the direction of propagation. The phase of the given wave is

$$\phi \;=\; 1.6\times10^{3}\,x \;+\; 48\times10^{10}\,t.$$

In a term of the form $$\sin(kx \;-\;\omega t)$$ the wave travels toward the positive x-axis, whereas for $$\sin(kx \;+\;\omega t)$$ the wave moves in the negative x-direction. Here the sign is positive, so the wave propagates toward the negative x-axis, i.e.

$$\vec k \;=\; -\hat i.$$

Next we use the right-hand rule for the cross product $$\vec E \times \vec B.$$ To obtain a vector pointing along $$-\hat i,$$ the cross product of $$\vec E$$ and $$\vec B=\;\hat j$$ must satisfy

$$\vec E \times \hat j \;=\; -\hat i.$$

Recalling the basic cyclic relation $$\hat i \times \hat j = \hat k,\; \hat j \times \hat k = \hat i,\; \hat k \times \hat i = \hat j,$$ we see that

$$\hat k \times \hat j \;=\; -\hat i.$$

Hence $$\vec E$$ must be along $$+\hat k.$$ (If it were along $$-\hat k,$$ the cross product would point toward $$+\hat i,$$ contradicting the actual propagation direction.)

Now we find the magnitude of the electric field. In free space the magnitudes of the fields in an EM wave are linked by the speed of light $$c$$ through the relation

$$E_0 \;=\; c\,B_0.$$

Here

$$B_0 \;=\; 3 \times 10^{-8}\,\text T,\qquad c \;=\; 3 \times 10^{8}\,\text{m s}^{-1}.$$

Substituting, we get

$$E_0 \;=\; \bigl(3 \times 10^{8}\bigr)\,\bigl(3 \times 10^{-8}\bigr)\;=\;9\;\text{V m}^{-1}.$$

The electric field therefore has amplitude $$9\;\text{V m}^{-1}$$, varies with the same phase $$\bigl(1.6\times10^{3}x + 48\times10^{10}t\bigr),$$ and points along $$+\hat k.$$ Putting everything together, the electric field is

$$\vec E \;=\; 9 \,\sin\!\bigl(1.6\times10^{3}\,x \;+\; 48\times10^{10}\,t\bigr)\,\hat k\;\frac{\text V}{\text m}.$$

Comparing with the options supplied, this matches Option B.

Hence, the correct answer is Option B.