jmc

The area of the shaded region is $(24s{)}^2$. To find the area of the large square, we note that there is a $d$-inch border between each of the $23$ pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of $23+2=25$ times the length of the border, i.e. $25d$. Adding this to the total length of the consecutive squares, which is $24s$, the side length of the large square is $(24s+25d)$, yielding the equation $\frac{(24s{)}^2}{(24s+25d{)}^2}=\frac{64}{100}$. Taking the square root of both sides (and using the fact that lengths are non-negative) gives $\frac{24s}{24s+25d}=\frac{8}{10}=\frac{4}{5}$, and cross-multiplying now gives $120s=96s+100d\Rightarrow 24s=100d\Rightarrow \frac{d}{s}=\frac{24}{100}=\boxed{\frac{6}{25}}$. Note: Once we obtain $\frac{24s}{24s+25d}=\frac{4}{5},$ to ease computation, we may take the reciprocal of both sides to yield $\frac{24s+25d}{24s}=1+\frac{25d}{24s}=\frac{5}{4},$ so $\frac{25d}{24s}=\frac{1}{4}.$ Multiplying both sides by $\frac{24}{25}$ yields the same answer as before.