Given below are the quantum numbers for $4$ electrons.

A. $\mathrm{n}=3,\mathrm{l}=2,{\mathrm{m}}_1=1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

B. $\mathrm{n}=4,\mathrm{l}=1,{\mathrm{m}}_1=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

C. $\mathrm{n}=4,\mathrm{l}=2,{\mathrm{m}}_1=-2,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$

D. $\mathrm{n}=3,\mathrm{l}=1,{\mathrm{m}}_1=-1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

The correct order of increasing energy is

Energy order of sub-shell decided by $\left(\mathrm{n}+\mathrm{l}\right)$ value.

Greater the $\left(\mathrm{n}+\mathrm{l}\right)$ value of a sub-shell greater is its energy.

If two sub-shells have same $\left(\mathrm{n}+\mathrm{l}\right)$ values the orbital with greater principal quantum number has greater energy

$\mathrm{A}\Rightarrow 3 \mathrm{d}\Rightarrow \left(\mathrm{n}+\mathrm{l}\right)=3+2=5$

$\mathrm{B}\Rightarrow 4\mathrm{p}\Rightarrow \mathrm{n}+\mathrm{l}=4+1=5$

$\mathrm{C}\Rightarrow 4 \mathrm{d}\Rightarrow \mathrm{n}+\mathrm{l}=4+2=6$

$\mathrm{D}\Rightarrow 3\mathrm{s}\Rightarrow \left(\mathrm{n}+\mathrm{l}\right)=3+0=3$

$\mathrm{D}<\mathrm{A}<\mathrm{B}<\mathrm{C}$