It is possible to estimate how far away you need to be from a
lightning strike while immersed in water and be reasonably safe.

The typical current in a lightning strike is about 30kA with a worst
case of about 300kA. The resistivity of seawater is 0.1 to 0.5 Ohm.m,
but the resistivity of lake waters varies from 100 to 400 Ohm.m.

Assume the lightning strike is a point source of current spreading
into a uniformly resistive hemisphere. There will be some distance at
which the potential difference between two points separated by say 2m
is at a level considered to be safe. A "safe" voltage is probably
somewhere in the region of 10 to 50V.

From "The Art and Science of Lightning Protection" by Uman, the
voltage between two points at distances a and b from the lightning
strike is

Vab = rho.I(1/a - 1/b)/2.pi

So in seawater with 0.5 Ohm.m resistivity for a 30kA lightning strike
at a distance of 20 to 22m the peak voltage drop will be about 11V.

However, in a 400 Ohm.m lake at 20m from the strike, the voltage drop
across 2m of water will be about 9kV.

At a range of 100m, in the 0.5 Ohm.m seawater the voltage drop will be
0.5V whereas in the 400 Ohm.m lake it will be about 370V.

I would get out of the lake!

John