Work done W = T × (change in surface area). The radius of each small droplet is r = R/2. Initial area = $4\pi R^2$. Final area = $8 \times (4\pi r^2) = 8 \times 4\pi (R/2)^2 = 8\pi R^2$. Change in area = $8\pi R^2 - 4\pi R^2 = 4\pi R^2$. Work done = $4\pi R^2 T$.