The decreasing order of energy for the electrons represented by the following sets of quantum numbers is 

$I.n=4,l=0,m=0,s=\pm 1/2, II.n=3,l=1,m=1,s=-1/2$

$III.n=3,l=2,m=0,s=+1/2, IV.n=3,l=0,m=0,s=-1/2$   

Smaller the value of $(n+l),$ smaller the energy. If two or more sub-orbits have same values of $(n+l),$ sub-orbits with lower values of $n$ has lower energy. The $(n+l)$ values of the given options are as follows

$I.n=4,l=0,n+l=4$

$II.n=3,l=1,n+l=4$

$III.n=3,l=2,n+l=5$

$IV.n=3,l=0,n+l=3$

Among I & II two II is having least ‘n’ value

Correct order is $3>1>2>4$