A resonance tube 1.0 m long has an inside diameter of 3.8 cm…

The wavelength. $\lambda = \dfrac{v}{f} = \dfrac{336\ \text{m/s}}{320\ \text{Hz}} = 1.05\ \text{m} = 105\ \text{cm}$.

The end correction. $e = 0.3 \times 3.8\ \text{cm} = 1.14\ \text{cm}$.

(a) $l_1 = \dfrac{105\ \text{cm}}{4} - 1.14\ \text{cm} = 26.25\ \text{cm} - 1.14\ \text{cm} = 25.1\ \text{cm}$.

(b) $l_2 = \dfrac{3 \times 105\ \text{cm}}{4} - 1.14\ \text{cm} = 78.75\ \text{cm} - 1.14\ \text{cm} = 77.6\ \text{cm}$.

(c) No. The third needs $l_3 = \dfrac{5 \times 105\ \text{cm}}{4} - 1.14\ \text{cm} = 130.1\ \text{cm}$, longer than the tube.