The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ  pipe is 110 Hz. The lengths of the pipes are

For closed organ pipe${\mathrm{f}}_{\mathrm{c}}=(2\mathrm{n}-1)\frac{\mathrm{v}}{4{\mathrm{L}}_{\mathrm{c}}}\text{ with }\mathrm{n}=1,2,\dots$
While for open organ pipe ${\mathrm{f}}_0=\mathrm{n}\frac{\mathrm{v}}{2{\mathrm{L}}_{\circ }}\text{ with }\mathrm{n}=1,2,\dots$
According to given problem
${\mathrm{f}}_{\mathrm{C}}=\frac{\mathrm{V}}{4{\mathrm{L}}_{\mathrm{c}}}=110\text{ i.e., }{\mathrm{L}}_{\mathrm{c}}=\frac{330}{4\times 110}=0.75\mathrm{m}$
Furthermore
$\begin{array}{l}\frac{3\mathrm{v}}{4{\mathrm{L}}_{\mathrm{c}}}~\frac{2\mathrm{v}}{2{\mathrm{L}}_{\circ }}=2.2\text{ or }\frac{330}{{\mathrm{L}}_{\circ }}=330\mp 2.2\left(\text{ as }\frac{3\mathrm{V}}{4{\mathrm{L}}_{\mathrm{c}}}=3{\mathrm{f}}_{\mathrm{C}}=330\right)\\ \text{ i.e., }{\mathrm{L}}_{\circ }=\frac{330}{330\mp 2.2}\Rightarrow {\mathrm{L}}_{\circ }=1.0067\mathrm{m}\text{ or }0.9937\mathrm{m}\end{array}$