A water drop is divided into 8 equal droplets. The pressure difference between inner and outer sides of the big drop

Suppose, R = radius of water drop and r = radius of droplets

$\frac{4}{3}{\mathrm{\pi R}}^3 = 8\times \frac{4}{3}{\mathrm{\pi r}}^3$

[Since, volume remains constant]

$\mathrm{r} = \frac{\mathrm{R}}{2}$

Since, excess pressure inside drop = $\frac{2\mathrm{T}}{\mathrm{R}}$

[T-surface tension, R-radius]

Pressure difference between inner and outer surface of big drop will be half of that for smaller droplet.