Can we call this Reuleaux Square?

I'll do it all in terms of $n$ . First of all, it is important to note that $\angle HDI = \angle KEH = \angle JFK = \angle IGJ = 30^o$ . Denoting $\overline {JK}$ as $x$ , by cosine law:

$\large \displaystyle x^2 = n^2 + n^2 - 2n^2 \cdot \cos(30^o)$

$\color{#20A900} \boxed{ \large \displaystyle x^2 = n^2 \cdot (2 - \sqrt{3}) }$

Area $HIJK$ is the area of a square of side $x$ plus 4 areas of circular segments, with angle $30^o$ and radius $n$ :

$\large \displaystyle A_{HIJK} = x^2 + 4 \cdot \left ( \frac{\pi n^2}{12} - A_{\Delta DHI} \right)$

Calculating $A_{\Delta DHI}$ by sine law and substituting $x^2$ :

$\large \displaystyle A_{HIJK} = n^2 \cdot (2 - \sqrt{3}) + 4 \cdot \left (\frac{\pi n^2}{12} - \frac{n^2}{4} \right )$

$\color{#20A900} \boxed{ \large \displaystyle A_{HIJK} = n^2 \cdot \left (1 + \frac{\pi}{3} - \sqrt{3} \right ) }$

now, to calculate $A_{DKHFIJ}$ , i.e., the area of the slit-looking part:

$\large \displaystyle A_{DKHFIJ} = n^2 - 2\cdot \left (n^2 - \frac{\pi n^2}{4} \right )$

$\color{#20A900} \boxed{ \large \displaystyle A_{DKHFIJ} = n^2 \left ( \frac{\pi}{2} - 1 \right )}$

The area of each corner of the slit will be:

$\large \displaystyle A_{DJK} = \frac12 \left (A_{DKHFIJ} - A_{HIJK} \right )$

$\large \displaystyle A_{DJK} = \frac12 \left [ n^2 \left ( \frac{\pi}{2} - 1 \right ) - n^2 \cdot \left (1 + \frac{\pi}{3} - \sqrt{3} \right ) \right ]$

$\color{#20A900} \boxed{ \large \displaystyle A_{DJK} = \frac12 n^2 \cdot \left ( \frac{\pi}{6} - 2 + \sqrt{3} \right ) }$

The whole purple area will then be:

$\large \displaystyle A_{PURPLE} = A_{HIJK} + 4 \cdot A_{DJK}$

$\large \displaystyle A_{PURPLE} = n^2 \cdot \left (1 + \frac{\pi}{3} - \sqrt{3} \right ) + 4 \cdot \frac12 n^2 \cdot \left ( \frac{\pi}{6} - 2 + \sqrt{3} \right )$

$\color{#20A900} \boxed{ \large \displaystyle A_{PURPLE} = n^2 \cdot \left ( \frac{2 \pi}{3} + \sqrt{3} - 3 \right ) }$

White area will be:

$\large \displaystyle A_{WHITE} = n^2 - A_{PURPLE}$

$\large \displaystyle A_{WHITE} = n^2 - n^2 \cdot \left ( \frac{2 \pi}{3} + \sqrt{3} - 3 \right )$

$\color{#20A900} \boxed{ \large \displaystyle A_{WHITE} = n^2 \left ( 4 - \frac{2 \pi}{3} - \sqrt{3}\right )}$

For $n = 4$ this leads to:

$\large \displaystyle A_{WHITE} = 16 \left ( 4 - \frac{2 \pi}{3} - \sqrt{3} \right )$

$\large \displaystyle A_{WHITE} = 32 \left ( 2 - \frac13 \pi - \frac12 \sqrt{3} \right )$

Which would lead to $a = 32, b = 2, c = 3, d = 2, e = 3, \boxed{a+b+c+d+e=42}$

However, this is not the correct answer according to the problem. I am assuming that $a$ , $b$ , $c$ , $d$ and $e$ must be all different. Re-writing $A_{WHITE}$ leads to:

$\color{#3D99F6} \boxed { \large \displaystyle A_{WHITE} = 64\left ( 1 - \frac16 \pi - \frac14 \sqrt{3} \right ) }$

So:

$\color{#3D99F6} \large \displaystyle a = 64, b = 1, c = 6, d = 4, e = 3, \boxed{\large \displaystyle a+b+c+d+e=78}$

The area of the required region is equal to $8$ times the area of the white region in the above figure

From the above figure it is clear that ,

$\text{Area of the white region}=\text{Area of yellow region}-\text{(Area of blue region+Area of red region)}$

Now $\Delta \text{DEH}$ is equilateral as all sides equals $a$

$\implies\angle HDE=60^{\circ}$

$\implies \angle GDH=30^{\circ} \hspace{4mm}\color{#3D99F6}\small \angle GDE=90^{\circ}\text{ as DEFG is a square}$

$\text{Area of yellow region}=\dfrac{\text{Area of square}}{2}=\dfrac{a^2}{2}$

$\text{Area of red region}=\text{Area of sector DHG}=a^2\cdot \dfrac{\pi}{12} \hspace{4mm}\color{#3D99F6}\small \text{Area of sector}=r^2\dfrac{\theta}{2}\text{ where r is the radius and } \theta \text{ the angle of the sector in radian}$

$\text{Area of the blue region}=\dfrac{\text{Area of }\Delta DEH}{2}=\dfrac{\sqrt{3}}{8}\cdot a^2\hspace{4mm}\color{#3D99F6}\small \text{Area of equilateral triangle with side a}=\dfrac{\sqrt{3}}{4}\cdot a^2$

Thus we have,

$\begin{aligned}\text{Area of white region}&=\dfrac{a^2}{2}-\left(a^2\cdot \dfrac{\pi}{12}+\dfrac{\sqrt{3}}{8}\cdot a^2\right)\\ &=\dfrac{a^2}{2}\left(1-\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\right)\end{aligned}$

$\begin{aligned}\text{Area of required region}&=8\times\text{Area of white region}\\\\ &=\dfrac{8a^2}{2}\left(1-\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\right)\end{aligned}$

$a=4$ in the above problem ,

$\begin{aligned}\implies\text{Area of required region}&=64\left(1-\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\right)\\ &=64-\dfrac{32\pi}{3}-16\sqrt{3}\\ \implies a+b+c+d+e&=64+32+3+16+3=\color{#EC7300}\boxed{\color{#333333}118}\end{aligned}$