Step-by-step Solution:

1. Analyze the given equation: We are given the equation $\overline z {z^3} + z{\overline z ^3} = 350$, where $z = x + iy$ and $x, y$ are integers.

2. Simplify the equation: We can factor out the common term $z\overline z$ from the left-hand side of the equation. $z\overline z (z^2 + {\overline z}^2) = 350$ We know that $z\overline z = |z|^2$. For $z = x+iy$, this is $x^2+y^2$. So the equation becomes: $|z|^2 (z^2 + {\overline z}^2) = 350$

3. Express the terms in x and y:

   • $|z|^2 = x^2+y^2$
   • To find $z^2 + {\overline z}^2$, we can use the identity $z^2 + {\overline z}^2 = (z+\overline z)^2 - 2z\overline z$. We know $z+\overline z = (x+iy) + (x-iy) = 2x$. So, $z^2 + {\overline z}^2 = (2x)^2 - 2|z|^2 = 4x^2 - 2(x^2+y^2) = 2x^2 - 2y^2 = 2(x^2 - y^2)$.

4. Substitute back into the simplified equation: Substituting the expressions for $|z|^2$ and $z^2 + {\overline z}^2$ into our simplified equation: $(x^2+y^2) [2(x^2-y^2)] = 350$ $(x^2+y^2)(x^2-y^2) = 175$

5. Solve for the integer values of x and y: Let $A = x^2$ and $B = y^2$. Since $x$ and $y$ are integers, $A$ and $B$ must be non-negative integers (specifically, perfect squares). The equation becomes: $(A+B)(A-B) = 175$ We need to find pairs of factors of 175. The factors of 175 are 1, 5, 7, 25, 35, 175. Let $A+B = u$ and $A-B = v$, where $uv=175$. Since $y^2 = B \ge 0$, we have $A+B \ge A-B$, which means $u \ge v$. Also, for $A = (u+v)/2$ and $B=(u-v)/2$ to be integers, $u$ and $v$ must have the same parity. Since 175 is odd, both $u$ and $v$ must be odd, which is true for all factors of 175.

   Let's check the possible pairs $(u, v)$:

   • Case 1: $u=175, v=1$ $A+B = 175$ and $A-B = 1$. Adding them: $2A = 176 \implies A = 88$. So $x^2 = 88$. This is not possible for an integer $x$.

   • Case 2: $u=35, v=5$ $A+B = 35$ and $A-B = 5$. Adding them: $2A = 40 \implies A = 20$. So $x^2 = 20$. This is not possible for an integer $x$.

   • Case 3: $u=25, v=7$ $A+B = 25$ and $A-B = 7$. Adding them: $2A = 32 \implies A = 16$. So $x^2 = 16 \implies x = \pm 4$. Subtracting them: $2B = 18 \implies B = 9$. So $y^2 = 9 \implies y = \pm 3$. This case gives integer solutions for $x$ and $y$.

6. Identify the roots of the equation: The integer solutions for $(x, y)$ are $(\pm 4, \pm 3)$. This gives us four complex numbers $z=x+iy$ which are the roots of the equation:

   • $z_1 = 4 + 3i$
   • $z_2 = 4 - 3i$
   • $z_3 = -4 + 3i$
   • $z_4 = -4 - 3i$

7. Calculate the area of the rectangle: The vertices of the rectangle in the Argand plane are the points corresponding to these roots: $(4, 3), (4, -3), (-4, 3),$ and $(-4, -3)$. The length of the rectangle is the distance between the points $(4,3)$ and $(-4,3)$, which is $|4 - (-4)| = 8$. The width of the rectangle is the distance between the points $(4,3)$ and $(4,-3)$, which is $|3 - (-3)| = 6$. The area of the rectangle is length $\times$ width. $\text{Area} = 8 \times 6 = 48$

8. Conclusion: The area of the rectangle is 48. This corresponds to option A.