$\text { Let the length of the side of each of the five similar squares be ' } x \text { ' each. }$


Now the square formed by joining the centre's of the four similar squares has side equal to the diagonal of one of the identical square i.e. $$\mathrm{x} \sqrt{2}$$ (Diagonal in a square is $$\sqrt{2}$$ times the side)

So, the area of each of the five identical squares and that of the square formed by joining the centers of the small squares is $$x^2 \&(x \sqrt{2})^2=2 x^2$$ respectively. (The area of square is (side) $${ }^2$$ ).

Finally the area of the new square formed is $$\frac{2 x^2-x^2}{x^2} \times 100=100 \%$$ more than that of each of the smaller (identical) squares.

Note: - As the area of the new square is 2 times that of the smaller squares, it is obviously $100 \%$ more.

Hence the answer is (E).