Hint: Recall total internal reflection.

Step 1: Find the angle of incidence.
The actual depth of the bulb in water, ${\mathrm{d}}_{\mathrm{l}}$= 80 cm = 0.8 m
Refractive Index of water, $\mu$= 1.33

Where,
i = Angle of incidence
r = Angle of refraction = 90°
Using Snell’s law,

$\begin{array}{l}\mu =\frac{\sin r}{\sin i}\\ 1.33=\frac{\sin {90}^{\circ }}{\sin i}\\ \Rightarrow i={\sin }^{-1}\left(\frac{1}{1.33}\right)=48.75°\end{array}$

Step 2: Find the radius of the circle.
Since the bulb is a point source, the emergent light can be considered as a circle of radius, R=$\frac{AC}{2}=OA$
Using the given figure,
The actual depth of the bulb in water, d_l = 80 cm = 0.8 m

$\tan i=\frac{OA}{OB}=\frac{R}{d_{\mathrm{l}}}$
$\therefore R=\tan 48.{75}^{\circ }\times 0.8=0.91m$

$$Step 3: Find the area of the circle.
Area of the circle = \(\pi r^{2}=\pi (0.91)^{2}=2.60\) m²
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.60 $m^2$.