Answer
Volume of the bubble at the surface of the lake is $V=2.249 \text{ mL}$
$d_1=10.9\text{ mm}$
$d_2=16\text{ mm}$
Work Step by Step
Using
$P_1V_1=P_2V_2$
$3.46\times0.650 = $1$\times$$V_2$
$V_2=2.249\text{ ml}$
Since the bubble is spherical, volume of bubble is:
$V_1 = 4/3\times$$\pi$$\times r^3$, $d_1=2r_1$
where $d_1$ is the diameter of bubble of volume $V_1$.
Solve for $r_1$:
$r_1=\sqrt[3]{\frac{3V_1}{4\pi}}=\sqrt[3]{\frac{3(0.650)\times 10^{-6}}{4\pi}}\approx 0.00545\text{ m}$
$d_1=2(0.00545)=0.0109\text{ m}$
In the same way we determine $d_2$:
$r_2=\sqrt[3]{\frac{3V_2}{4\pi}}=\sqrt[3]{\frac{3(0.2.25)\times 10^{-6}}{4\pi}}\approx 0.0080\text{ m}$
$d_2=2(0.0080)=0.016\text{ m}$
