Answer
$ ( -\color{black}{\bf 256}\;\hat i +\color{black}{\bf 138}\;\hat j -\color{black}{\bf 177}\;\hat k )\;\rm W/m^2$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field is always perpendicular to the magnetic field.
So the dot product at any point must be zero,
$$\vec E\cdot \vec B=0$$
Hence,
$$(200\;\hat i+300\;\hat j-50\;\hat k)\cdot B_0(7.3\;\hat i-7.3\;\hat j+a\;\hat k)\times 10^{-6}=0$$
Recalling that $\hat i\cdot \hat i=1$, and $\hat i\cdot \hat j=0$, so
$$(200\times 7.3 )-(300\times 7.3) -(50a) =0$$
Thus,
$$a=\color{red}{\bf -14.6}$$
Recalling that
$$E=cB$$
Hence,
$$\sqrt{(200\;\hat i)^2+(300\;\hat j)^2+(-50\;\hat k)^2}=c\sqrt{ B_0^2[(7.3\;\hat i)^2+(-7.3\;\hat j)^2+(-14.6\;\hat k)^2]\times (10^{-6})^2}$$
$$\sqrt{(200)^2+(300)^2+(50)^2}=cB_0\sqrt{ [(7.3)^2+(7.3)^2+(14.6)^2]\times (10^{-6})^2}$$
Thus,
$$B_0=\dfrac{\sqrt{(200)^2+(300)^2+(50)^2}}{c\sqrt{ [(7.3)^2+(7.3)^2+(14.6)^2]\times (10^{-6})^2}}$$
Plug the known;
$$B_0=\dfrac{\sqrt{(200)^2+(300)^2+(50)^2}}{(3\times 10^8)\sqrt{ [(7.3)^2+(7.3)^2+(14.6)^2]\times (10^{-6})^2}}$$
$$B_0=\color{red}{\bf 6.78\times 10^{-2}}\;\rm T$$
$$\color{blue}{\bf [b]}$$
The Poynting vector is given by
$$\vec S=\dfrac{1}{\mu_0}\left[\vec E\times \vec B\right]\tag 1$$
Let's find $\vec E\times \vec B$;
$$\vec E\times \vec B=(200\;\hat i+300\;\hat j-50\;\hat k)\times (7.3\;\hat i-7.3\;\hat j-14.6\;\hat k)(0.0678\times 10^{-6}) $$
$$\vec E\times \vec B=\\\left[
(200\;\hat i\times 7.3\;\hat i)+(200\;\hat i\times -7.3\;\hat j)
+(200\;\hat i\times -14.6\;\hat k)\\+(300\;\hat j\times 7.3\;\hat i)
+(300\;\hat j\times -7.3\;\hat j)+(300\;\hat j\times -14.6\;\hat k)\\
+(-50\;\hat k\times 7.3\;\hat i)+(-50\;\hat k\times -7.3\;\hat j)+(-50\;\hat k\times -14.6\;\hat k)\right]\\(0.0678\times 10^{-6})
$$
Recalling that $\hat i\times \hat i=\hat j\times \hat j=\hat k\times \hat k=0$;
$$\vec E\times \vec B=0-1460\;\hat k+2920\;\hat j-2190\;\hat k+0-4380\;\hat i-365\;\hat j-365\;\hat i+0$$
$$\vec E\times \vec B=(-4745\;\hat i +2555\;\hat j -3285\;\hat k)(0.0678\times 10^{-6})$$
Plug into (1);
$$\vec S=\dfrac{(0.0678\times 10^{-6})}{(4\pi\times 10^{-7})}\left[ -4745\;\hat i +2555\;\hat j -3285\;\hat k \right] $$
$$\vec S=0.054\left[ -4745\;\hat i +2555\;\hat j -3285\;\hat k \right]$$
$$\vec S= ( -\color{red}{\bf 256}\;\hat i +\color{red}{\bf 138}\;\hat j -\color{red}{\bf 177}\;\hat k )\;\rm W/m^2$$
