Wavelengths of different radiations are given below:
$\lambda(A) = 300 \ nm$
$\lambda(B) = 300 \ \mu m$
$\lambda(C) = 3 \ nm$
$\lambda(D) = 30 \ \mathring{A}$
Arrange these radiations in the increasing order of their energies.

(B < A < C = D) The energy of a radiation is given by $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
First,convert all wavelengths to meters:
$\lambda(A) = 300 \ nm = 300 \times 10^{-9} \ m = 3 \times 10^{-7} \ m$
$\lambda(B) = 300 \ \mu m = 300 \times 10^{-6} \ m = 3 \times 10^{-4} \ m$
$\lambda(C) = 3 \ nm = 3 \times 10^{-9} \ m$
$\lambda(D) = 30 \ \mathring{A} = 30 \times 10^{-10} \ m = 3 \times 10^{-9} \ m$
Comparing the wavelengths: $\lambda(B) > \lambda(A) > \lambda(C) = \lambda(D)$.
Since energy is inversely proportional to wavelength,the increasing order of energy is $B < A < C = D$.