1. We need to identify the acid with the lowest $pK_a$.
   Lower $pK_a$ means stronger acid.

2. Given acids:

   • A: $CH_3COOH$ = acetic acid
   • B: $HCOOH$ = formic acid
   • C: $(CH_3)_2COOH$ (intended as a more alkyl-substituted carboxylic acid)
   • D: $CH_3CH_2COOH$ = propionic acid

3. Key principle:
   In carboxylic acids, electron-donating alkyl groups increase electron density on the carboxylate ion and destabilize the conjugate base, making the acid weaker. Hence, more alkyl substitution means higher $pK_a$.

4. Compare the acids:

   • $HCOOH$ has no alkyl group attached.
   • $CH_3COOH$ has one methyl group.
   • $CH_3CH_2COOH$ has an ethyl group, which is electron-donating.
   • $(CH_3)_2COOH$ is even more alkyl-substituted/electron-donating.

   Therefore, acidity order is approximately: $HCOOH > CH_3COOH > CH_3CH_2COOH > (CH_3)_2COOH$

   So the corresponding $pK_a$ order is: $HCOOH < CH_3COOH < CH_3CH_2COOH < (CH_3)_2COOH$

5. Hence, the lowest $pK_a$ is for $HCOOH$.

6. Option check:

   • A: $CH_3COOH$ — not lowest
   • B: $HCOOH$ — lowest $pK_a$
   • C: $(CH_3)_2COOH$ — not lowest
   • D: $CH_3CH_2COOH$ — not lowest

Final answer: B