Answer
$\approx 4.4\;\rm cm$
Work Step by Step
Since the author told us to assume that the air is an ideal fluid, we can use Bernoulli’s equation.
We chose point 1 at the left-wider pipe and point 2 at the same height (on the same horizontal line) at the right-narrower pipe. This means that $y_2-y_1=0$.
$$P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
Hence,
$$P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
$$P_1-P_2= \frac{1}{2}\rho v_2^2-\frac{1}{2}\rho v_1^2+\rho gy_2-\rho gy_1$$
$$P_1-P_2= \frac{1}{2}\rho (v_2^2- v_1^2)+\overbrace{\rho g(y_2- y_1)}^{=0}$$
$$\overbrace{P_1-P_2}^{h\rho_{Hg} g}= \frac{1}{2}\rho (v_2^2- v_1^2) $$
Noting that the pressure difference $P_1-P_2=h\rho_{Hg}g$ since the pressure above the mercury surface in the left tube is $P_1$ and the pressure above the mercury surface in the right tube is $P_2$.
$$ h\rho_{Hg} g = \frac{1}{2}\rho (v_2^2- v_1^2)\tag 1 $$
Now we need to find $v_1$ and $v_2$, and we know that the volume rate is given by
$$Q=A_1v_1=A_2v_2$$
Hence,
$$v_1=\dfrac{Q}{A_1}=\dfrac{Q}{\pi r_1^2}=\dfrac{Q}{\pi\left[\frac{d_1}{2}\right]^2}$$
$$v_1=\dfrac{4Q}{\pi d_1^2}\tag 2$$
And by the same approach,
$$v_2 =\dfrac{4Q}{\pi d_2^2}\tag 3$$
Plugging (2) and (3) into (1);
$$ h\rho_{Hg} g = \frac{1}{2}\rho \left(\left[\dfrac{4Q}{\pi d_2^2}\right]^2- \left[\dfrac{4Q}{\pi d_1^2}\right]^2\right) $$
$$ h\rho_{Hg} g = \frac{16Q^2}{2\pi^2}\rho \left( \dfrac{1}{ d_2^4} - \dfrac{1}{ d_1^4} \right) $$
$$ h = \frac{8Q^2\rho }{ \pi^2\rho_{Hg} g}\left( \dfrac{1}{ d_2^4} - \dfrac{1}{ d_1^4} \right) $$
Plugging the known and remember to convert the units to SI units.
$$ h = \frac{8(1200\times 10^{-6})^2(1.28) }{ \pi^2(13600)(9.8)}\left( \dfrac{1}{ (4\times 10^{-3})^4} - \dfrac{1}{ (2\times 10^{-2})^4} \right) $$
$$h=0.0437 \;\rm m\approx \color{red}{\bf 4.4}\;\rm cm$$
