1. Given magnetic field

   $\vec B = 2\times 10^{-8}\sin\left(0.5\times 10^3 x + 1.5\times 10^{11} t\right)\,\hat j\ \text{T}$

   So, the amplitude of magnetic field is

   $B_0 = 2\times 10^{-8}\ \text{T}$

2. Relation between electric and magnetic fields in an electromagnetic wave

   For a plane electromagnetic wave in vacuum,

   $E_0 = cB_0$

   where

   $c = 3\times 10^8\ \text{m/s}$

   Therefore,

   $E_0 = (3\times 10^8)(2\times 10^{-8}) = 6\ \text{V/m}$

3. Direction of propagation

   The phase is

   $0.5\times 10^3 x + 1.5\times 10^{11} t$

   A wave of the form

   $\sin(kx-\omega t)$

   travels in the $+x$ direction, while

   $\sin(kx+\omega t)$

   travels in the $-x$ direction.

   Hence this wave propagates along the negative $x$-axis.

4. Determine direction of electric field

   For an electromagnetic wave,

   $\vec E \times \vec B = \text{direction of propagation}$

   Here,

   $\vec B \parallel \hat j$

   and propagation is along

   $-\hat i$

   We need $\vec E$ such that

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

   Now,

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

   Therefore,

   $\vec E \parallel \hat k$

   i.e. electric field is along the $z$-axis.

5. Check options

   • A: $6\,\text{V/m}$ along $x$-axis $\rightarrow$ wrong direction
   • B: $3\,\text{V/m}$ along $z$-axis $\rightarrow$ wrong magnitude
   • C: $6\,\text{V/m}$ along $z$-axis $\rightarrow$ correct
   • D: $2\times 10^{-8}\,\text{V/m}$ along $z$-axis $\rightarrow$ wrong magnitude

Therefore, the correct option is C.