Given electric field

$\vec E = E_0\hat i\cos(kz)\cos(\omega t)$

So the electric field is along the $x$-direction.

Use Maxwell’s equation

For electromagnetic waves,

$\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$

Since

$E_x = E_0\cos(kz)\cos(\omega t), \quad E_y=0, \quad E_z=0$

the curl is

$\nabla \times \vec E = \left(\frac{\partial E_x}{\partial z}\right)\hat j$

with a minus sign from the determinant:

$\nabla \times \vec E = \left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\hat i + \left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)(-\hat j) + \left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)\hat k$

More directly,

$\nabla\times \vec E = -\frac{\partial E_x}{\partial z}\hat j$

Now,

$\frac{\partial E_x}{\partial z} = -kE_0\sin(kz)\cos(\omega t)$

hence

$\nabla \times \vec E = kE_0\sin(kz)\cos(\omega t)\hat j$

Relate to magnetic field

From

$\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$

we get

$-\frac{\partial \vec B}{\partial t} = kE_0\sin(kz)\cos(\omega t)\hat j$

so

$\frac{\partial \vec B}{\partial t} = -kE_0\sin(kz)\cos(\omega t)\hat j$

Integrate with respect to time

$\vec B = -\frac{kE_0}{\omega}\sin(kz)\sin(\omega t)\hat j$

Using

$\frac{\omega}{k}=c \quad \Rightarrow \quad \frac{k}{\omega}=\frac{1}{c}$

therefore

$\vec B = -\frac{E_0}{c}\sin(kz)\sin(\omega t)\hat j$

Compare with options

The magnetic field must be along $\hat j$, with dependence $\sin(kz)\sin(\omega t)$. The sign depends on propagation convention/superposition form, but among the given options only A has the correct functional dependence and direction.

Hence the best matching option is:

$\boxed{\vec B = \frac{E_0}{c}\hat j\sin(kz)\sin(\omega t)}$

i.e. Option A.