As shown in the figure, the cross-sectional area of the wider tube...

Let the difference in water levels be $h$.

The additional pressure exerted by the boy on the wider tube is balanced by the pressure due to the extra height of the water column in the narrower tube.

Equating the pressures at the same horizontal level in both tubes:

$\dfrac{mg}{A} = \rho g h$

$h = \dfrac{m}{\rho A}$

Given:
Mass of the boy, $m = 45 \text{ kg} = 45000 \text{ g}$
Cross-sectional area, $A = 900 \text{ cm}^2$
Density of water, $\rho = 1 \text{ g/cm}^3$

Substituting the values:

$h = \dfrac{45000}{1 \times 900} \text{ cm}$

$h = 50 \text{ cm}$