The strongest acid amongst the following compounds is:

For carboxylic acids, acidity is governed mainly by the stability of the conjugate base $$RCOO^-$$.
The more the negative charge on oxygen is dispersed (stabilised), the stronger the parent acid is.

Electron-withdrawing groups (-I effect) stabilise the conjugate base, while electron-releasing alkyl groups ( +I effect) destabilise it. The -I effect dies out rapidly with distance, so a substituent placed α (one carbon) to the -COOH group has a far larger impact than if it is β or γ.

Let us inspect each option.

Option A $$HCOOH$$ (formic acid)
No alkyl group, hence no +I destabilisation. pKa ≈ 3.75.

Option B $$CH_3CH_2CH(Cl)CO_2H$$ (α-chlorobutyric acid)
Chlorine, an electronegative atom, sits on the α-carbon. Its strong -I effect greatly stabilises $$CH_3CH_2CH(Cl)COO^-$$. Although two CH2 groups are present, their +I effect is weaker and is farther away than the α-chlorine. Net result: highest stabilisation and pKa ≈ 2.8.

Option C $$ClCH_2CH_2CH_2COOH$$ (γ-chlorobutyric acid)
Here chlorine is three carbons away (γ-position). Its -I effect is attenuated by distance, while three CH2 groups contribute considerable +I destabilisation. Hence it is less acidic than formic acid. pKa ≈ 4.5.

Option D $$CH_3COOH$$ (acetic acid)
A single CH3 group exerts a +I effect, giving the least stable conjugate base of the four. pKa ≈ 4.76.

Ordering the acids from strongest to weakest:
$$\text{α-chlorobutyric acid (B)} \; \gt \; \text{formic acid (A)} \; \gt \; \text{γ-chlorobutyric acid (C)} \; \gt \; \text{acetic acid (D)}$$

Therefore, the strongest acid is Option B which is: $$CH_3CH_2CH(Cl)CO_2H$$.