What is the buoyant force on an object that weighs 340 n and is floating on a lake? if you could explain the answer that would be nice.

You just said that the object is "floating".  

(As soon as you said that, a picture of a duck flashed through my mind.  But then I knew right away that the duck could not be an accurate representation of the situation you're describing.  340 N would be some duck ... about 76 pounds ... and that duck would have been caught and eaten a long time ago. I mean ... what could a 76-pound duck do ?  Could it fly away ?  Could it run away ? ?  Not likely.)

So it's not a duck, but whatever it is, it's just sitting there on the water, floating.  What's important is that it's not accelerating up or down.  THAT tells us that the vertical forces on it are balanced so that there's NO NET vertical force on it at all.

What are the vertical forces on it ?  There's gravity, pulling it DOWN with a force of 340 N, and there's buoyancy, pushing it UP.  The SUM of those two forces must be zero ... otherwise the object would be accelerating up or down.

It's not.  So (gravity) + (buoyancy) must add up to zero.

The buoyant force on the object is 340 N UPward.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft