Gauge pressure due to liquid at a depth of $5 \mathrm{~cm}$
$$
p=\rho g h
$$
Density, $\rho=1000 \mathrm{~kg} / \mathrm{m}^3, h=5 \mathrm{~cm}=0.05 \mathrm{~m}$,
$$
\begin{aligned}
& g=10 \mathrm{~m} / \mathrm{s}^2 \\
& p=1000 \times 10 \times 0.05 \\
& p=500 \mathrm{~Pa}
\end{aligned}
$$
$\Rightarrow$ Excess pressure inside the air bubble w.r.t. outside due to surface tension is $p^{\prime}=\frac{2 S}{R}$
Surface tension, $S=0 \mathrm{l} \mathrm{N} / \mathrm{m}$
Radius, $R=2 \mathrm{~mm}=0.002 \mathrm{~m}$
$$
p^{\prime}=\frac{2 \times 01}{0.002}=100 \mathrm{~Pa}
$$
Now, total excess pressure inside the bubble w.r.t surface of liquid, i.e.
$$
\begin{aligned}
\Rightarrow \quad p_{\text {total }} & =p+p^{\prime} \\
& =500+100=600 \mathrm{~Pa}
\end{aligned}
$$