For a plane electromagnetic wave, the magnetic field at a point x and time t is:
$$\vec{B}(x, t) = [1.2 \times 10^{-7}\sin(0.5 \times 10^3 x + 1.5 \times 10^{11}t)\hat{k}]\,\text{T}$$.
The instantaneous electric field $$\vec{E}$$ corresponding to $$\vec{B}$$ is:

For any plane electromagnetic wave travelling in free space, the oscillating electric field $$\vec E$$ and magnetic field $$\vec B$$ are always perpendicular to each other as well as to the direction of propagation. The two fundamental relations that connect them are:

1. Magnitude relation  $$E_0 = c\,B_0$$  where  $$c = 3 \times 10^8\;{\rm m\,s^{-1}}$$ is the speed of light.
2. Direction relation  $$\vec k \times \vec E = \dfrac{\omega}{c}\,\hat k_B = \omega\mu_0\vec H$$, or more simply, in vacuum the vector product $$\vec E \times \vec B$$ points along the wave-vector $$\vec k$$, i.e. the direction in which the wave energy propagates.

The given magnetic field is

$$\vec{B}(x,t) = \bigl[\,1.2 \times 10^{-7}\,\sin(0.5 \times 10^{3}\,x + 1.5 \times 10^{11}\,t)\,\hat k\bigr]\;{\rm T}.$$

Comparing this expression with the standard harmonic form $$\sin(kx \pm \omega t)$$, we identify

$$k = 0.5 \times 10^{3}\;{\rm m^{-1}}, \qquad \omega = 1.5 \times 10^{11}\;{\rm s^{-1}}.$$

Because the argument of the sine is $$kx + \omega t$$ (with a plus sign), the condition for constant phase $$kx + \omega t = {\rm constant}$$ gives

$$k\,\dfrac{dx}{dt} + \omega = 0 \;\;\Longrightarrow\;\; \dfrac{dx}{dt} = -\,\dfrac{\omega}{k}.$$

Since $$dx/dt$$ is negative, the electromagnetic wave travels along the negative x-direction, i.e. along $$-\,\hat i$$.

The magnetic field vector is along $$+\,\hat k$$. To obtain the correct direction of $$\vec E$$ we use the right-hand rule so that $$\vec E \times \vec B$$ points along the direction of propagation $$-\,\hat i$$:

$$\vec E \times \vec B \;=\; -\,\hat i.$$ Because $$\vec B$$ is along $$+\,\hat k$$, the only unit vector that satisfies this cross-product is $$-\,\hat j$$ since

$$( -\,\hat j ) \times ( +\,\hat k ) = -\,(\hat j \times \hat k) = -\,\hat i.$$

Hence the electric field must be directed along $$-\,\hat j$$.

Next we calculate its amplitude. From the magnitude relation $$E_0 = c B_0$$ we substitute the numerical values:

$$E_0 = (3 \times 10^{8}\,{\rm m\,s^{-1}})\,(1.2 \times 10^{-7}\,{\rm T}) = 3.6 \times 10^{1}\,{\rm V\,m^{-1}} = 36\;{\rm V\,m^{-1}}.$$

Finally we write the complete expression for $$\vec E(x,t)$$, using the same space-time dependence as $$\vec B$$ (the fields oscillate in phase):

$$\vec E(x,t) = \bigl[\, -\,36 \,\sin(0.5 \times 10^{3}\,x + 1.5 \times 10^{11}\,t)\,\hat j \bigr]\;\dfrac{{\rm V}}{{\rm m}}.$$

Comparing with the given options, this matches Option A.

Hence, the correct answer is Option A.