= Solution
For identical spherical drops, total oil volume and interfacial area satisfy
$$
V_o=N\frac{4\pi r^3}{3},\qquad A=N4\pi r^2=\frac{3V_o}{r},\qquad E_s=\gamma_{ow}A.
$$
Taking $10\ \mathrm{vol}\%$ to mean the oil fraction of the final <emulsion>, $V_o/(V_o+V_w)=0.10$ gives $V_o=V_w/9=1.111\times10^{-4}\ \mathrm{m^3}$. Thus $A=333.3\ \mathrm{m^2}$ and \b[$E_s=16.7\ \mathrm J$]. This corresponds to about $2.65\times10^{13}$ drops.
There is a wording convention worth making explicit: if the oil addition instead means $0.10$ times the initial water volume, then $V_o=100\ \mathrm{mL}$, $A=300\ \mathrm{m^2}$ and \b[$E_s=15.0\ \mathrm J$]. These differ only in the stated volume-fraction convention. The calculation uses the oil–water interface and neglects the container and air interfaces.
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