Answer
$7.6\;\rm mm$
Work Step by Step
Since we have two mediums, we need to use the formula of
$$\dfrac{n_1}{s}+\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}$$
where $n_1=n_{\rm air}=1$, and $n_2=n_{\rm aqueous~ humor}=1.34$.
The author asks for the radius of curvature so we need to solve for $R$.
$$R=\dfrac{n_2-n_1}{\left[\dfrac{n_1}{s}+\dfrac{n_2}{s'}\right]}$$
Now we can assume that $s=\infty$ when we are dealing with parallel rays coming. And since we need to focus these ryas in the humor, $s'=f$.
$$R=\dfrac{n_2-n_1}{\left[\dfrac{n_1}{\infty}+\dfrac{n_2}{f}\right]}$$
Plugging the known;
$$R=\dfrac{1.34-1}{\left[0+\dfrac{1.34}{3}\right]}=\bf 0.76\;\rm cm$$
$$R=\color{red}{\bf 7.6}\;\rm mm$$
