Energy associated with second Bohr's orbit is
$
\begin{aligned}
E & =-13.6 \mathrm{Z}^2 / n^2 \\
E & =-13.6 \times 3^2 / 2^2 \\
E & =-30.6 \mathrm{eV} \\
& =-30.6 \times 1.6 \times 10^{-19} \\
& =4.905 \times 10^{-18} \mathrm{~J}
\end{aligned}
$
Radius of second Bohr's orbit in $\mathrm{Li}^{+}$is given by
$
\begin{aligned}
R & =0.529 n^2 / Z \\
R & =0.529 \times 2^2 / 3 \\
R & =0.7053 Å=0.0705 \mathrm{~nm}
\end{aligned}
$
Hence, the correct option is (4)