Answer
$24\;\rm cm$
Work Step by Step
Let's assume that the temperature of the air is constant during this process of compressing it by pouring the mercury. And let's assume that the air sealed is obeying the ideal gas law.
Thus,
$$\dfrac{P_1V_1}{ \color{red}{\bf\not} T_1}=\dfrac{P_2V_2}{ \color{red}{\bf\not} T_2}$$
$$P_1V_1=P_2V_2\tag 1$$
Initially, the volume of the air was the volume of the whole U-tube and its pressure was the atmospheric pressure of 1 atm.
Finally, the air is compressed by a distance of $L$ which is the length of the mercury column, and its final pressure is the atmospheric pressure plus the pressure due to the mercury column.
So that,
$$P_a( \color{red}{\bf\not} Al)=(P_A+L\rho g)( \color{red}{\bf\not} A[l-L])$$
where $l$ is the total length of the U-tube, $A$ is its cross-sectional area, $L$ is the length of the mercury column, $\rho$ is the mercury density.
$$P_al=(P_A+L\rho g) [l-L] $$
Solving for $L$;
$$P_al=P_al-P_aL+lL\rho g-L^2\rho g$$
$$0= -P_a \color{red}{\bf\not} L+l \color{red}{\bf\not} L\rho g-L^{ \color{red}{\bf\not} 2}\rho g$$
$$L\rho g=l\rho g-P_a$$
$$L=\dfrac{l\rho g-P_a}{\rho g}$$
$$L=l -\dfrac{P_a}{\rho g}$$
Plugging the known;
$$L=1.0 -\dfrac{(1.013\times 10^5)}{(13600)(9.8)}$$
$$L=\color{red}{\bf 0.24}\;\rm m$$
