Pressure due to this height of $Hg$, acts downward on the trapped air
In this position, the trapped air is under a total pressure $P_1$ $P_1=P_{\text{amm }}+$ pressure due to $15cm$ of $Hg$
$P_1=H+15cm$ (since $P_{\text{atm }}$ and $P$ due to $15cmHg$ both act $P=H+15cm$ downward on the trapped air)
In this position, the air is under a total pressure
$P_2=P_{2\text{ atm }}-15cm$ $P_2=(H-15)cm$
$P_2=(H-15)cm$ In this position, the trapped air is under a total pressure $P_3$
$=P_{\text{stm }}=H$y Boyle's law, $PV=$ constant
i.e. $P_1{V}_1=P_2{V}_2$
where $V=A\times l$
$P_1{A}_1{l}_1=P_2\times A_2{l}_2$
Since the tube is uniform, $A_1=A_2$
So that
$P_1{l}_1=P_2{l}_2$
$(H+15)\times 10=(H-15)\times 15$
$H+15=(H-15)\times 1.5$
$H+15=1.5H-22.5$
$H=75cm$
$P_{atm}=75cm$