The fundamental frequency of a closed organ pipe is equal to the...

$\dfrac{\lambda}{4} = {L}_{1}2\left(\dfrac{\lambda}{2}\right) = {L}_{2}$

$\text{As}, v = f\lambda ⇒ {f}_{2} = \dfrac{2v}{2{L}_{2}}$

$⇒ v = {f}_{1}\left(4{L}_{1}\right) ⇒ {f}_{2} = \dfrac{v}{{L}_{2}}$

$⇒ {f}_{1} = \dfrac{v}{4{L}_{1}}$

As per question, ${f}_{1} = {f}_{2}$

$\dfrac{v}{4{L}_{1}} = \dfrac{v}{{L}_{2}}$

$⇒ {L}_{2} = 4{L}_{1}$

$⇒ 60 = 4 \times {L}_{1}$

$⇒ {L}_{1} = 15cm$