**Answer:** $A$

Let the roots be $\alpha_1, \alpha_2, \alpha_3, \alpha_4,$ and $\alpha_5$

Now,

We know that the roots of the coefficients are the real numbers and two imaginary roots are given.

Therefore with this hint, we know other two roots as well, which are nothing but just the conjugate of these two imaginary roots. So, total $4$ roots are known out of $5$

Now, for any polynomial.

$$(x+\alpha_1)(x + \alpha_2)(x+\alpha_3)(x+\alpha_4)(x+\alpha_5) = 0$$

$$\therefore \; x^5+ x^4(\alpha_1 + \alpha_2+\alpha_3+\alpha_4+\alpha_5)+x^3(\alpha_1\alpha_2+\alpha_2\alpha_3+\dots) + x^2(\alpha_1.\alpha_2.\alpha_3 + \alpha_2.\alpha_3.\alpha_4+\dots) + x^1(\alpha_1\alpha_2\alpha_3\alpha_4 + \alpha_2\alpha_3\alpha_4\alpha_5+\dots) + x^0(\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5) = 0$$

So, product of roots is given by:

$$\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5 = 4 \implies (1+2\iota)(1-2\iota)(3-2\iota)(3+2\iota) \color{blue}{\alpha_5} = 4$$

$$\implies \color{blue}{\alpha_5} = \frac{4}{65}$$

Now, Sum of of roots is given by:

$$\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5 = -a$$

$$\implies (1+2\iota)+(1-2i)+(3+2\iota)+(3-2\iota)+\frac{4}{65} = \color{red} {-a}$$

$$\implies \color {red}{a} = \color {red}{-\frac{524}{65}}$$

$\therefore \;A$ is the correct answer.