A charge of $$4\mu C$$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

Total charge available: $$Q = 4\,\mu C$$. This charge is split into two parts: one part is $$q$$ and the other is $$Q-q = 4 - q$$, with $$0 \le q \le 4$$.

The two charges are kept at a fixed separation $$r$$. Coulomb’s law gives the magnitude of the electrostatic force between them as

$$F = k\,\frac{q\,(4-q)}{r^{2}}$$

where $$k = 9 \times 10^{9}\,{\rm N\,m^{2}\,C^{-2}}$$ is Coulomb’s constant.
Since $$k$$ and $$r$$ are constants, maximising $$F$$ is the same as maximising the product

$$f(q) = q\,(4-q)$$

Expand $$f(q)$$:

$$f(q) = 4q - q^{2}$$

This is a concave-down parabola in $$q$$. The maximum of a quadratic $$-q^{2}+4q$$ occurs at its vertex.
For a quadratic $$ax^{2}+bx+c$$ with $$a \lt 0$$, the vertex is at $$x = -\frac{b}{2a}$$.
Here, $$a = -1$$ and $$b = 4$$, so

$$q_{\text{max}} = -\frac{4}{2(-1)} = 2$$

Thus the first charge should be $$q = 2\,\mu C$$ and the second charge is

$$(4 - q)\,\mu C = (4 - 2)\,\mu C = 2\,\mu C$$

Therefore, to obtain the maximum possible electrostatic force, the charge should be divided equally:

Case of maximum force: $$2\,\mu C$$ and $$2\,\mu C$$

Option B is correct.