1. Given magnetic field

   $B_y = \left(3.5\times 10^{-7}\right)\sin\left(1.5\times 10^3 x + 0.5\times 10^{11} t\right)\,\text{T}$

   So the magnetic field is along the $y$-direction.

2. Direction of propagation

   The phase is of the form $kx + \omega t$ A wave written as $\sin(kx+\omega t)$ travels in the negative $x$-direction.

3. Direction of electric field

   For an electromagnetic wave, $\vec E \perp \vec B \perp \text{direction of propagation}$

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

   Here propagation is along $-\hat x$, and $\vec B \parallel \hat y$

   We need $\vec E$ such that $\vec E \times \hat y = -\hat x$

   Since $\hat z \times \hat y = -\hat x,$ the electric field must be along the $z$-direction.

   Hence, $E = E_z$

4. Magnitude relation between $E$ and $B$

   In free space, $E_0 = cB_0$

   Given $B_0 = 3.5\times 10^{-7}\,\text{T}, \qquad c = 3\times 10^8\,\text{m/s}$

   Therefore, $E_0 = (3\times 10^8)(3.5\times 10^{-7}) = 10.5\times 10^1 = 105\,\text{V/m}$

5. Write the electric field

   The electric and magnetic fields are in phase, so

   $E_z = 105\sin\left(1.5\times 10^3 x + 0.5\times 10^{11} t\right)\,\text{V m}^{-1}$

6. Compare with options

   • A: $E_z=105\sin\left(1.5\times 10^3 x+0.5\times 10^{11} t\right)\,\text{V m}^{-1}$ ✅
   • B: wrong direction and wrong magnitude
   • C: wrong direction and wrong magnitude
   • D: correct direction but wrong magnitude

Therefore, the correct option is A.