To solve this problem, we need to use the formula for the radius of the $n$-th orbit of a hydrogen atom.$\\$The formula for the radius of the $n$-th orbit is given by:$\\ r_n = n^2 \cdot r_1 \\$where:$\\$• $r_n$ is the radius of the $n$-th orbit,$\\$• $n$ is the principal quantum number,$\\$• $r_1$ is the radius of the first orbit (innermost orbit).$\\$Given:$\\ r_1 = 5.3 \times 10^{-11}$ m$\\$We need to find the radius of the third orbit $(n = 3). \\$Substitute the values into the formula:$\\ r_3 = 3^2 \cdot r_1 \\ r_3 = 9 \cdot 5.3 \times 10^{-11}$ m$\\$Calculate $r_3: \\ r_3 = 47.7 \times 10^{-11}$ m$\\$Convert $r_3$ to Angstroms $(\overset{\circ}{A}): \\ 1$ m$= 10^{10} \overset{\circ}{A} \\ r_3 = 47.7 \times 10^{-11}$ m$\times 10^{10} \overset{\circ}{A}/$m$\\ r_3 = 4.77 \overset{\circ}{A} \\$Therefore, the radius of the third allowed orbit of the hydrogen atom is $4.77 \overset{\circ}{A}. \\$This corresponds to Option 4.