Let l = 3x, b = 2x and h = x

Area of 4 walls = 2(lh + bh) = 2(3x^{2} + 2x^{2}) = 10x^{2}

New dimensions are as follows,

l' = 2l = 6x, b' = $\frac{b}{2}$ = x and h' = $\frac{h}{2}$ = 0.5x

Area of 4 walls = 2(l'h' + b'h') = 2(6x * 0.5x + x * 0.5x) = 7x^{2}

So we can see that area has decreased.

% decrease = $\frac{10x^{2} - 7x^{2}}{10x^{2}}$ * 100 = 30%

**Option 5 is correct**