Solution:
Differentiating the given integral with respect to $a$ gives $f(g(a)) = e^{-2a}$. Now
$$x \frac{d[\ln (f(g(x)))]}{dx} = x \frac{f'(g(x)) g'(x)}{f(g(x))} = \frac{g'(f(x)) f'(x)}{g(f(x))} = \frac{d[\ln (g(f(x)))]}{dx}$$
where the second equals sign follows from the given. Since $\ln (f(g(x))) = -2x$, we have $-x^2 + C = \ln (g(f(x)))$, so $g(f(x)) = K e^{-x^2}$. It follows that $K = 1$ and $g(f(4)) = e^{-16}$.