24) C

Let side of first square = A = 1
Let side of second square = B
Let side of third square = C
Let side of fourth square = D

Using Pythagoras theorem,

Second Square
B² = (¹⁄₃)² + (²⁄₃)² = ¹⁄₉ + ⁴⁄₉ = ⁵⁄₉
B = √⁵⁄₉ = √5 
                   3

Third square
C² = (¹⁄₃ x √5/3) ² + (²⁄₃ x √5/3) ²  = (√5 / 9)² + (2√5 / 9)² = ⁵⁄₈₁ + ²⁰⁄₈₁ = ²⁵⁄₈₁
C = √²⁵⁄₈₁ = ⁵⁄₉

Fourth square
D² = (⅓ x ⁵⁄₉ ) ² + (²⁄₃ x ⁵⁄₉) ²  = (⁵⁄₂₇)² + (¹⁰⁄₂₇)² = ²⁵⁄₇₂₉ + ¹⁰⁰⁄₇₂₉ = ¹²⁵⁄₇₂₉
D = √¹²⁵⁄₇₂₉

Area of fourth square = D x D = D² = (√¹²⁵⁄₇₂₉) = ¹²⁵⁄₇₂₉