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# Hint

The forward scatting is proportional to the sum:
$$I(0)\propto\Sigma_i(\Delta\mathrm{SLD}_i V_i)^2$$
where $\Delta\mathrm{SLD}_i = \mathrm{SLD}_i - \mathrm{SLD}_\mathrm{solvent}$ is the contrast (or excess scattering length density) of subunit $i$, and $V_i$ is the volume of that subunit. A core-shell particle consists of two subunits (core and the shell), defined by regions with approximately constant contrast.

In this problem, $\Delta\mathrm{SLD}_\mathrm{core} = -\Delta\mathrm{SLD}_\mathrm{shell}$, so the volumes of the core and shell must be equal in order to get $I(0)=0$.

The volume of the core is $V_\mathrm{core} = \frac{4\pi}{3}R_\mathrm{core}^3$, and the shell volume is $V_\mathrm{shell} = \frac{4\pi}{3}R_\mathrm{outer}^3- V_\mathrm{core}$, so
$$\frac{4\pi}{3}R_\mathrm{core}^3 = \frac{4\pi}{3}R_\mathrm{outer}^3 - \frac{4\pi}{3}R_\mathrm{core}^3$$
which gives a relation between the radii:
$$2R_\mathrm{core}^3 = R_\mathrm{outer}^3$$