A tunning fork of frequency 340Hz resonates in the fundamental mode...

We know that, $v = \lambdaf$, then, $\lambda = \dfrac{v}{f} = \dfrac{340m{s}^{- 1}}{340Hz} = \dfrac{340m{s}^{- 1}}{340{s}^{- 1}} = 1m$

First resonating length, ${l}_{1} = \dfrac{\lambda}{4} = \dfrac{100}{4} = 25cm$

Second resonating length, ${l}_{2} = \dfrac{3\lambda}{4} = \dfrac{3 \times 100}{4} = 75cm$

Third resonating length, ${l}_{3} = \dfrac{5\lambda}{4} = \dfrac{5 \times 100}{4} = 125cm$

So third resonance is not possible since the length of the tube is $125cm$

Thus, minimum height of water necessary for resonance $= 125 - 75 = 50cm$