The correct option is 2d.

Let us understand the step-by-step reasoning behind this answer.

According to Coulomb's Law, the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $d$ in vacuum is given by the formula:

$$F=k·\frac{q_1{q}_2}{d^2}$$

where $k$ is Coulomb's constant.

Now, let the new charges be $q_1^{\prime }=2q_1$ and $q_2^{\prime }=2q_2$. Let the new distance between them to keep the force unchanged be $d^{\prime }$.

The new force $F^{\prime }$ is expressed as:

$$F^{\prime }=k·\frac{q_1^{\prime }{q}_2^{\prime }}{{\left(d^{\prime }\right)}^2}$$

Substituting the values of the new charges:

$$F^{\prime }=k·\frac{\left(2q_1\right)\left(2q_2\right)}{{\left(d^{\prime }\right)}^2}=k·\frac{4q_1{q}_2}{{\left(d^{\prime }\right)}^2}$$

We are given that the force remains unchanged, meaning $F^{\prime }=F$. Therefore, we set the two equations equal to each other:

$$k·\frac{4q_1{q}_2}{{\left(d^{\prime }\right)}^2}=k·\frac{q_1{q}_2}{d^2}$$

By canceling the common terms $k$, $q_1$, and $q_2$ from both sides, we get:

$$\frac{4}{{\left(d^{\prime }\right)}^2}=\frac{1}{d^2}$$

Cross-multiplying to solve for $d^{\prime }$ gives:

$${\left(d^{\prime }\right)}^2=4d^2$$

Taking the square root on both sides:

$$d^{\prime }=2d$$

Thus, to keep the force unchanged, the distance between the charges must be doubled to 2d.