The electric field of an electromagnetic wave in free space is
$$\vec{E} = 57 \cos \left[ 7.5 \times 10^{6} t - 5 \times 10^{-3} (3x + 4y) \right] (4\hat{i} - 3\hat{j}) N / C$$. The associated magnetic field in Tesla is

We are given the electric field $$ \vec{E} = 57 \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](4\hat{i} - 3\hat{j}) \text{ N/C} $$. The direction of propagation is along the wave vector $$\vec{k}$$, which from the phase is $$ \hat{k} = \frac{3\hat{i} + 4\hat{j}}{5} $$. Since the direction of $$\vec{E}$$ is along $$(4\hat{i} - 3\hat{j})$$, we verify that it is perpendicular to $$\hat{k}$$ by computing $$(3\hat{i} + 4\hat{j}) \cdot (4\hat{i} - 3\hat{j}) = 12 - 12 = 0$$, confirming orthogonality.

For an electromagnetic wave, the magnetic field $$\vec{B}$$ is given by $$ \vec{B} = \frac{\hat{k} \times \vec{E}}{c} $$. Now we compute the cross product:

$$ \hat{k} \times (4\hat{i} - 3\hat{j}) = \frac{1}{5}(3\hat{i} + 4\hat{j}) \times (4\hat{i} - 3\hat{j}) = \frac{1}{5}\left[3(-3)(\hat{i} \times \hat{j}) + 4(4)(\hat{j} \times \hat{i})\right] = \frac{1}{5}\left[-9\hat{k} - 16\hat{k}\right] = \frac{-25\hat{k}}{5} = -5\hat{k}. $$

Substituting back into the expression for $$\vec{B}$$ gives $$ \vec{B} = \frac{57}{c} \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](-5\hat{k}) $$. Therefore, using $$c = 3 \times 10^{8}\text{ m/s}$$, we have $$\vec{B} = -\frac{57}{3 \times 10^{8}} \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](5\hat{k})$$ T $$.$$

The correct answer is Option 3.