Answer
$28.04\;\rm cm$
Work Step by Step
As we see in the figure below, the water flows from point 1 to point 2.
We can use Bernoulli’s equation to find the pressure at each point.
So,
$$\color{red}{\bf\not}P_1+\frac{1}{2}\rho v_1^2+\rho g y_1=\color{red}{\bf\not}P_2+\frac{1}{2}\rho v_2^2+\rho g y_2$$
We can see that both $P_1$ and $P_2$ are atmospheric pressure.
$$ \rho g y_1-\rho g y_2= \frac{1}{2}\rho v_2^2- \frac{1}{2}\rho v_1^2$$
Noting that $y_1-t_2=h$, if we chose $y_2=0$
$$ \color{red}{\bf\not}\rho g h= \frac{1}{2}\color{red}{\bf\not}\rho (v_2^2- v_1^2)$$
$$ g h= \frac{1}{2} (v_2^2- v_1^2)\tag1 $$
Now we need to find $v_1$ and $v_2$. We can find them by the given rate of flow and the continuity equation, where
$$Q=A_1v_1=A_2v_2$$
Hence,
$$v_1=\dfrac{Q}{A_1}=\dfrac{Q}{\pi r_1^2}=\dfrac{Q}{\pi\left[\frac{d_1}{2}\right]^2}$$
And by the same approach,
$$v_2=\dfrac{Q}{A_2}=\dfrac{Q}{\pi r_2^2}=\dfrac{Q}{\pi\left[\frac{d_2}{2}\right]^2}$$
Plugging these two velocitites into (2);
$$ g h= \frac{1}{2}\left[\left(\dfrac{Q}{\pi\left[\frac{d_2}{2}\right]^2}\right)^2- \left(\dfrac{Q}{\pi\left[\frac{d_1}{2}\right]^2}\right)^2\right]$$
$$ h= \dfrac{1}{2g }\left[\dfrac{16Q^2}{\pi^2 d_2^4 }- \dfrac{16Q^2}{\pi^2 d_1^4 }\right]= \dfrac{16Q^2}{2\pi^2g }\left[\dfrac{1}{ d_2^4 }- \dfrac{1}{ d_1^4 }\right]$$
$$ h = \dfrac{8Q^2}{ \pi^2g }\left[\dfrac{1}{ d_2^4 }- \dfrac{1}{ d_1^4 }\right]$$
Plugging the known and remember to convert the units to SI units.
$$ h = \dfrac{8(0.2\times 10^{-3})^2}{ \pi^2(9.8) }\left[\dfrac{1}{ (10\times 10^{-3})^4 }- \dfrac{1}{ (16\times 10^{-3})^4 }\right]$$
$$h=0.2804\;\rm m=\color{red}{\bf 28.04}\;\rm cm$$
