Explanation

(c)
Give that,
Bohr’s orbit of hydrogen atom ($n$)=2
Atomic number of hydrogen ($Z$)=1
By using
$\ \ \ \ \ \ \ \ \ \ \ r = \frac{0.529\ n^{2}}{Z}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{0.529 \times (2)^{2}}{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{0.529 \times 4}{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.116\ \mathring{\mathrm{A}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.2116\ nm$