Fork A gives its first resonance at 20.8 cm in a tube, and the second resonance is 43.0 cm further…

(a) $\lambda/2 = 43.0\ \text{cm}$, so $\lambda = 0.860\ \text{m}$. We want $f$, so $f = \dfrac{v}{\lambda} = \dfrac{344\ \text{m/s}}{0.860\ \text{m}} = 400\ \text{Hz}$.

(b) $l_1 + e = \lambda/4 = 21.5\ \text{cm}$, so $e = 21.5\ \text{cm} - 20.8\ \text{cm} = 0.7\ \text{cm}$.

(c) B is $400 + 6$ or $400 - 6$ Hz. Its first resonance is lower down the tube, so the column is longer and its wavelength is longer. So B has the lower frequency: $394\ \text{Hz}$.

(d) $\dfrac{\lambda_B}{4} = \dfrac{344\ \text{m/s}}{4 \times 394\ \text{Hz}} = 21.83\ \text{cm}$, so $l_1 = 21.83\ \text{cm} - 0.7\ \text{cm} = 21.1\ \text{cm}$.