Step 1. Given Data:
Initial velocity, $u=10 \mathrm{~m} / \mathrm{s}$
Angle of projection, $\theta=60^{\circ}$
Radius of curvature at $t=1 s$ is $R$
Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}$
Step 2. Finding the radius of curvature:
Resolving the velocity into $x$ and $y$ components, at any time $t$
$$
\begin{aligned}
& v_x=u \cos 60^{\circ}=10 \times \frac{1}{2} \\
& v_x=5 \mathrm{~m} / \mathrm{s} \\
& v_y=u \sin 60^{\circ}+a t=\frac{10 \sqrt{3}}{2}-10 t \\
& v_y=|5 \sqrt{3}-10 t|
\end{aligned}
$$



At $t=1 s$
$$
v_y=|5 \sqrt{3}-10|
$$

Angle made by body with respect to horizontal axis at $t=1 \mathrm{~s}$
$$
\begin{aligned}
& \tan \alpha=\left|\frac{v_y}{v_x}\right|=\left|\frac{5 \sqrt{3}-10}{5}\right|, \\
& \tan \alpha==|\sqrt{3}-2| \\
& \alpha=\tan ^{-1}(|\sqrt{3}-2|)=15^{\circ}
\end{aligned}
$$

The radius of curvature same as range at time $t$, which is given by
$$
\begin{aligned}
R & =\frac{v^2}{g \cos \alpha}=\frac{v_x^2+v_y^2}{g \cos \alpha} \\
R & =\frac{5^2+(5 \sqrt{3}-10)^2}{10 \times \cos 15^o} \\
R & =\frac{25+75+100-100 \sqrt{3}}{10 \times \cos 15^{\circ}} \\
R & =\frac{26.79}{9.65}=2.77 \\
R & =2.77 m \approx 2.8 m
\end{aligned}
$$

Hence, option $C$ is correct