Answer
$76\;\rm cm$
Work Step by Step
According to the hint, we can use Boyle’s law
$$P_1V_1=P_2V_2$$
Now we know that the initial volume of the air was the whole cylinder.
We also know that the volume is given by $V=AL$ where $A$ is the cross-sectional area of the cylinder and $L$ is its length.
This means that the initial height of the air is $L$ and the final height of the air is $h$.
So,
$$P_1\color{red}{\bf\not}A L=P_2\color{red}{\bf\not}A h $$
Hence,
$$ P_1L=hP_2 $$
Noting that $P_1=P_a$, and that $L=1$ m
$$P_a=hP_2 \tag 1$$
The final pressure of the air is given by
$$P_2=P_a+h_{1}\rho_{1}g$$
Note that the mass of the piston is negligible, otherwise, we shall account for it.
where $h_1$ refers to the height of the mercury column and $\rho_1$ refers to the mercury density.
From the given figure, we can see that $ h_1=L-h=(1-h)$
So,
$$P_2=P_a+(1-h)\rho_{1}g$$
Plugging into (1);
$$P_a=h[P_a+(1-h)\rho_{1}g]$$
$$P_a-h P_a=(1-h)h\rho_{1}g $$
$$P_a\color{red}{\bf\not}(1-h) =\color{red}{\bf\not}(1-h)h\rho_{1}g $$
Thus,
$$h=\dfrac{P_a}{\rho_1g}$$
Plugging the known;
$$h=\dfrac{(1.013\times 10^5)}{(13600)(9.8)}=\bf 0.760\;\rm m$$
$$h=\color{red}{\bf 76}\;\rm cm$$
