jmc

If $9x^2+24x+a$ is the square of a binomial, then the binomial has the form $3x+b$ for some number $b$, because $(3x{)}^2=9x^2$. So, we compare $(3x+b{)}^2$ to $9x^2+24x+a$. Expanding $(3x+b{)}^2$ gives $$(3x+b{)}^2=(3x{)}^2+2(3x)(b)+b^2=9x^2+6bx+b^2.$$Equating the linear term of this to the linear term of $9x^2+24x+a$, we have $6bx=24x$, so $b=4$. Equating the constant term of $9x^2+6bx+b^2$ to that of $9x^2+24x+a$ gives us $a=b^2=\boxed{16}$.