Let a = the length of the side of the cube.

The perpendicular distance from F to the line segment LK is 10. Since FL = FK the perpendicular distance from F to LK is the height of the triangle FLK. Let F1 be the middle point of the segment LK.

LK is the middle line for the triangle ABD thus LK = (1/2) DB, DB is the diagonal for the square ABCD and from Pythagorean Theorem (DB)^2 = 2a^2 now we have

(LK)^2 = (DB/2)^2 = 2a^2/4 = (a^2)/2

also the triangle FKB is right triangle with legs KB = a/2 and FB^2 = 2a^2 and hypotenuses FK. Using the Pythagorean Theorem we can write for the hypotenuses

FK^2 = KB^2 + FB^2

FK^2 = (a/2)^2 + 2a^2 = a^2/4 + 2a^2 = 9a^2/4

The triangle FF1Kis right triangle and again using the Pythagorean Theorem we have

FF1^2 + F1K^2 = FK^2

F1K^2 = (LK/2)^2 = LK^2/4 = ((a^2)/2)/4 = a^2/8

FF1 = 10

FK^2 = 9a^2/4

if we plug these value in FF1^2 + F1K^2 = FK^2 we get

10^2 + a^2/8 = 9a^2/4

by solving we find

a = 6.86

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the volume of the cube is

V = a^3

V = 6.86^3

V = 323

The volume of the cube to the nearest integer is 323.