Answer
$544\;\rm nm$
Work Step by Step
As we see below, we have two reflected rays with one $\pi$-phase change.
This means that the path length difference for destructive interference is given by
$$2t=m\lambda_n=\dfrac{m\lambda}{n}\tag 1$$
and for constructive interference the path length difference is given by
$$2t=\left(m+\frac{1}{2}\right)\lambda_n=\dfrac{\left(m+\frac{1}{2}\right)\lambda}{n}\tag 2$$
And since we have a maximum reflected light for two wavelengths, we will use equation (2) to find the thickness of the film.
- For $\lambda_1=688\;\rm nm$,
$$t=\dfrac{\left(m_1+\frac{1}{2}\right)\lambda_1}{2n}\tag 3$$
- For $\lambda_2=491.4\;\rm nm$,
$$t=\dfrac{\left(m_2+\frac{1}{2}\right)\lambda_2}{2n}\tag 4$$
The left side of the two last formulas (3) and (4) is the same, so doe the right side.
Thus;
$$ \dfrac{\left(m_2+\frac{1}{2}\right)\lambda_2}{2n}=\dfrac{\left(m_1+\frac{1}{2}\right)\lambda_1}{2n} $$
$$ \left(m_2+\frac{1}{2}\right)\lambda_2 = \left(m_1+\frac{1}{2}\right)\lambda_1 $$
$$ \dfrac{\lambda_1}{\lambda_2} =\dfrac{\left(m_2+\frac{1}{2}\right)}{\left(m_1+\frac{1}{2}\right) } $$
Thus,
$$ \dfrac{\lambda_1}{\lambda_2} =\dfrac{\left(m_1+1+\frac{1}{2}\right)}{\left(m_1+\frac{1}{2}\right) } $$
Plugging the known
$$ \dfrac{688}{491.4}= \dfrac{\left(m_1 +\frac{3}{2}\right)}{\left(m_1+\frac{1}{2}\right) } $$
$$ \dfrac{688}{491.4}m_1+\dfrac{688}{491.4} \cdot \frac{1}{2}= \left(m_1 +\frac{3}{2}\right) $$
$$ \dfrac{688}{491.4} m_1 -m_1= \frac{3}{2}-\dfrac{688}{982.8} $$
Thus, $m_1=1.9999\approx 2$.
Plugging that into (3)
$$t=\dfrac{\left(2+\frac{1}{2}\right)\cdot 688}{2\cdot 1.58} =\color{red}{\bf 544}\;\rm nm$$