The radius of the 2 nd orbit of Li ^ 2 + is x . The expected radius...

Radius of orbit $( r ) = 0 . 529 \times \dfrac{{n}^{2}}{Z}$

Where, n = principal quantum number of orbit

Z = Atomic number

$( r{Li}^{2 +}{)}_{n = 2} = 0 . 529 \times \dfrac{( 2{)}^{2}}{3} = x - - - - - ( 1 )$

${\left(r{Be}^{3 +}\right)}_{n = 3} = 0 . 529 \times \dfrac{( 3{)}^{2}}{4} - - - - - ( 2 )$

Divide equation (1) by equation (2)

$\dfrac{{\left(r{Li}^{2 +}\right)}_{n = 2}}{{\left(r{Be}^{3 +}\right)}_{n = 3}} = \dfrac{0 . 529 \times \dfrac{( 2{)}^{2}}{3}}{0 . 529 \times \dfrac{( 3{)}^{2}}{4}}$

$⇒ \dfrac{x}{( r{Be}^{3 +}{)}_{n = 3}} = \dfrac{16}{27}$

$∴ {\left(r{Be}^{3 +}\right)}_{n = 3} = \dfrac{27x}{16}$