Answer
Using Kepler's Second Law we show that the time required to travel from $A'$ to $B'$ is
${t_{A'B'}} = \left( {\frac{1}{4} + \frac{e}{{2\pi }}} \right)T$
Work Step by Step
Recall that the area of the ellipse is $A = \pi ab$. So, by symmetry, the area of $OA'B'$ is ${A_1} = \frac{1}{4}\pi ab$.
Let us denote the position of the Sun as ${F_2}$. So, the area of the triangle ${F_2}OB'$ is ${A_2} = \frac{1}{2}bc$. Since, $e = \frac{c}{a}$, so, ${A_2} = \frac{1}{2}abe$.
Thus, the area swept out by the planet from $A'$ to $B'$ is
${A_{A'B'}} = \frac{1}{4}\pi ab + \frac{1}{2}abe = ab\left( {\frac{1}{4}\pi + \frac{1}{2}e} \right)$
Using Kepler's Second Law which states that the position vector pointing from the Sun to the planet sweeps out equal areas in equal times, we have
$\frac{{{A_{A'B'}}}}{A} = \frac{{{t_{A'B'}}}}{T}$
where ${t_{A'B'}}$ is the time required to travel from $A'$ to $B'$.
Therefore,
$\frac{{ab\left( {\frac{1}{4}\pi + \frac{1}{2}e} \right)}}{{\pi ab}} = \frac{{{t_{A'B'}}}}{T}$
${t_{A'B'}} = \left( {\frac{1}{4} + \frac{e}{{2\pi }}} \right)T$,
as required.
