If length, width and height are $l$, $w$ and $h$, then the new dimensions are $l \frac{100+p}{100}$, $w \frac{100+p}{100}$, and $h \frac{100-q}{100}$. Those fractions are imposing, so instead of the percentages $p$ and $q$, let's work with the fractions $x = \frac{p}{100}$ and $y = \frac{q}{100}$. In these terms the new dimensions are $l (1+x)$, $w (1+x)$, and $h (1-y)$.

The volume of the solid is supposed to be unchanged. That is,

$l (1+x) w (1+x) h (1-y) = lwh.$

The original dimensions cancel and we are left with the equation,

$(1+x)^2(1-y) = 1,$

in which $y$ is the unknown. Solving is easy enough:

Putting back the $p$'s and $q$' is not so nice:

The answer, believe it or not, is (e).