Let the radius of the big circle be $\text{R cm},$ and the radius of the small circle be $r \;\text{cm}.$

We can get from $\text{case M}$ and $\text{case N.}$

- $\text{R + R + R + R} = 1 \Rightarrow \text{4R} = 1 \Rightarrow {\color{Blue}{\boxed{\text{R} = \frac{1}{4}\;\text{cm}}}}$
- $r+r+r+r+r+r = 1 \Rightarrow 6r = 1 \Rightarrow {\color{Purple}{\boxed{r = \frac{1}{6}\;\text{cm}}}}$

Now, we can find the area of the circles in both the cases.

- $\text{Case M:}$
- Area of the four circles $ = 4 \times \pi \text{R}^{2}$

- $\text{Case N:}$
- Area of the nine circles $ = 9 \times \pi \text{r}^{2}$

The area of the square in both the cases $ = 1^{2} = 1$

Now, we can find the areas of unshaded regions.

- $\text{Case M:}$
- The area of unshaded regions $ = 1-4\pi\text{R}^{2} = 1 – 4\pi \left(\frac{1}{4}\right)^{2} = 1 – \frac{\pi}{4}$

- $\text{Case N:}$
- The area of unshaded regions $ = 1-9\pi\text{r}^{2} = 1-9\pi \left(\frac{1}{6}\right)^{2} = 1 – \frac{\pi}{4}$

$\therefore$ The ratio of the areas of unshaded regions of $\text{case M}$ to that of $\text{case N}$ is $1:1.$

Correct Answer $:\text{B}$

${\color{Magenta}{\textbf{PS:}}}$

- ${\color{Green}{\text{The area of square} = \text{(side)}^{2}}}$
- ${\color{Lime}{\text{The area of circle} = \pi \times (\text{radius})^{2}}}$