Sphere inside of a Sphere

If the inner sphere of dough has radius $r$ and when the outer shell of choclate has radius $R$ ... and we require the volume of dough & chocolate to be equal then \begin{eqnarray*} 2 \frac{4 \pi r^3}{3} = \frac{4 \pi R^3}{3}. \end{eqnarray*} So we require the ratio $R/r = \sqrt[3]{2}=1.2599 \cdots$. So a good approximation would be $ \frac{5}{4}$.

In other words if your ball of dough is $4$ units then when it is covered in chocolate make the radius increase to $5$ units.

Note that we rounded $1.2599$ down to $1.25$ so there will be slightly less chocolate ... so I would advise you to put an extra smidge of chocolate on in order to remedy this ... you can't put too much chocolate on your cookies!

Consider cookie as $3D$ object. So if corresponding $1D$ size of cookie + chocolate is $a > 1$ where $1D$ size of cookie is $1$, then $3D$ size of cookie with chocolate will be $a^3$.

Now, you have

$$ \frac{cookie + chocolate}{cookie} = \frac{a^3}{1} = 2 $$

Or

$$ a^3 = 2 $$

Therefore chocolate addition to diameter must be $\sqrt[3]{2} - 1$