A diver in water picks up a lead cube of side length 10 cm . How much force is needed to lift the cube? The density of lead is approximately 11 g/(cm^3) .

There are three forces acting on the lead cube:gravitational force with the magnitude Mg downwards,buoyant (Archimedes') force mg upwards,lifting force of unknown magnitude F upwards.
Here g is the gravity acceleration, M is the mass of the cube and m is the mass of water in the volume of the cube. Note that  M = rho_l V,  m = rho_w V,  V = a^3, where a is the side length, V is the volume of the cube, rho_l is the density of lead and rho_w approx1 g/(cm)^3 is the density of water.
If we find F_0 such that these forces will be balanced, then any lifting force magnitude greater than F_0 value will be suitable.
Considering the directions, the equation becomes
F_0 +rho_w a^3 g =rho_l a^3 g,  or
F_0 =rho_l a^3 g -rho_w a^3 g = (rho_l -rho_w) a^3 g.
We need to divide this by 1000 to obtain kilograms from grams. The numerical result is about  (11 - 1)*10^3*10/1000 = 100 (N).
Thus the answer is: at least 100 N force is needed to lift the cube.
http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html