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Let the height of the tower be $h$ and the total time taken to reach the ground be $t$ seconds.
Since the body falls freely from rest ($u = 0$):
$h = \frac{1}{2}gt^2$

The distance traveled in the last second (i.e., the $t$-th second) is:
$h_{\text{last}} = \frac{1}{2}g(2t - 1)$

According to the question:
$h_{\text{last}} = \frac{4}{9}h$
$\frac{1}{2}g(2t - 1) = \frac{4}{9}\left(\frac{1}{2}gt^2\right)$
$2t - 1 = \frac{4}{9}t^2$
$4t^2 - 18t + 9 = 0$

Solving for $t$ using the quadratic formula:
$t = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(4)(9)}}{2(4)} = \frac{18 \pm \sqrt{324 - 144}}{8} = \frac{18 \pm \sqrt{180}}{8} = \frac{9 \pm 3\sqrt{5}}{4}$

Since the fall takes at least 1 second ($t > 1$):
$t = \frac{9 + 3\sqrt{5}}{4} \approx 3.927 \text{ s}$
(The other root $t \approx 0.573\text{ s} < 1$, which is inadmissible).

Taking $g = 9.8 \text{ m/s}^2$:
$h = \frac{1}{2}gt^2 = \frac{1}{2} \times 9.8 \times (3.927)^2 \approx 75.56 \text{ m}$

(Note: If the fraction in the standard textbook problem is $\frac{5}{9}$ instead of $\frac{4}{9}$, then $5t^2 - 18t + 9 = 0 \implies (t - 3)(5t - 3) = 0 \implies t = 3\text{ s}$, which gives $h = \frac{1}{2} \times 9.8 \times 3^2 = 44.1\text{ m}$.)