The increasing order of the atomic radii of the following elements is:
(a) C
(b) O
(c) F
(d) Cl
(e) Br

We begin with the basic periodic trends for atomic (covalent) radius.

1. Along any period (that is, from left to right in the same row of the periodic table) the atomic number $$Z$$ increases but the electrons are being added to the same principal energy level. The effective nuclear charge $$Z_{\text{eff}}$$ therefore increases, pulling the electron cloud closer to the nucleus.
So, within a single period: $$\text{Atomic radius decreases from left to right.}$$

2. Down any group (that is, from top to bottom in the same column) a new principal energy level is added each time. Even though $$Z_{\text{eff}}$$ also rises a little, the increase in principal quantum number $$n$$ dominates, keeping the outer electrons farther from the nucleus.
So, within a group: $$\text{Atomic radius increases from top to bottom.}$$

Now we list the given elements with their periodic positions:

$$\begin{aligned} \text{C}&:& \text{Group 14, Period 2}\\ \text{O}&:& \text{Group 16, Period 2}\\ \text{F}&:& \text{Group 17, Period 2}\\ \text{Cl}&:& \text{Group 17, Period 3}\\ \text{Br}&:& \text{Group 17, Period 4} \end{aligned}$$

First, let us compare the three elements that lie in the same period (Period 2): $$\text{C, O, F}.$$ Moving from C to O to F we move left → right, so the radii must decrease:

$$r_{\text{F}} < r_{\text{O}} < r_{\text{C}}.$$

Next, consider the three halogens $$\text{F, Cl, Br}$$ which lie in the same group (Group 17). Going down the group the radius increases, hence

$$r_{\text{F}} < r_{\text{Cl}} < r_{\text{Br}}.$$

To place C among Cl and Br we recall that C is in Period 2 while Cl is in Period 3. Because an extra principal shell is present in Period 3, $$r_{\text{Cl}}$$ is larger than $$r_{\text{C}}.$$ Numerically, their covalent radii are approximately $$r_{\text{C}}\approx 77\,\text{pm},\; r_{\text{Cl}}\approx 99\,\text{pm}.$$ Thus

$$r_{\text{C}} < r_{\text{Cl}}.$$

Combining every inequality obtained, starting from the smallest value and moving upward:

$$r_{\text{F}} < r_{\text{O}} < r_{\text{C}} < r_{\text{Cl}} < r_{\text{Br}}.$$

Translating back to the element symbols given in the question: $$(c)\;F \; <\; (b)\;O \; <\; (a)\;C \; <\; (d)\;Cl \; <\; (e)\;Br.$$

This sequence matches exactly with Option C.

Hence, the correct answer is Option C.