The electric field midway between two charges can be found by calculating the electric fields due to each charge at that point and then adding them vectorially. In this case, we have a 300 mC charge at the origin and a -300 mC charge 1 m to the right of it. The midpoint between the two charges is located at (0.5, 0) m.

The magnitude of the electric field due to a point charge is given by Coulomb's law:
$$E = \frac{kq}{r^2}$$
where $k=9 \times 10^9 \text{ N m}^2/\text{C}^2$ is Coulomb's constant, $q$ is the charge, and $r$ is the distance from the charge to the point where the electric field is being calculated.

The electric field due to the 300 mC charge at the midpoint is:
$$E_1 = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2)(300 \times 10^{-3} \text{ C})}{(0.5 \text{ m})^2} = 10800 \times 10^6 \text{ N/C}=10.8\times 10^9\text{N/C}.$$

The electric field due to the -300 mC charge at the midpoint is:
$$E_2 = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2)(300 \times 10^{-3} \text{ C})}{(0.5\text{ m})^2} = 10.8 \times 10^9 \text{ N/C}$$

The magnitude of total electric field at the midpoint is the vector sum of these two fields:
$$E_{\text{total}} = E_1 + E_2= 21.6\times10^9 \text{ N/C}$$

The electric field emerges from a positive charge and converges towards a negative charge. As a result, at the midpoint, the electric field direction will align with the positive X-axis due to the influence of both the positive and negative charges.

Therefore, the answer is (A), i.e., the magnitude of the electric field is $21.6$ $\times$ 10$^9$ N/C, and to the right.