The quantum number of four electrons are given below:
I. $$n = 4, l = 2, m_l = -2, m_s = -1/2$$
II. $$n = 3, l = 2, m_l = 1, m_s = +1/2$$
III. $$n = 4, l = 1, m_l = 0, m_s = +1/2$$
IV. $$n = 3, l = 1, m_l = 1, m_s = -1/2$$
The correct order of their increasing energies will be:

First we recall the empirical (n + l) rule, also called the Madelung rule, which states: “For multi-electron atoms the sub-shell having smaller $$n + l$$ value possesses lower energy. If two sub-shells have the same $$n + l$$ value, the one with smaller principal quantum number $$n$$ is lower in energy.”

Now we evaluate the quantity $$n + l$$ for every given electron:

For electron I we have $$n = 4$$ and $$l = 2$$. So $$ n + l = 4 + 2 = 6. $$

For electron II we have $$n = 3$$ and $$l = 2$$. So $$ n + l = 3 + 2 = 5. $$

For electron III we have $$n = 4$$ and $$l = 1$$. So $$ n + l = 4 + 1 = 5. $$

For electron IV we have $$n = 3$$ and $$l = 1$$. So $$ n + l = 3 + 1 = 4. $$

Next we arrange the sub-shells in ascending order of their $$n + l$$ values:

Electron IV has $$n + l = 4$$, the smallest of all, so it is the lowest in energy.

Electrons II and III both have $$n + l = 5$$. For such a tie we compare their principal quantum numbers $$n$$. Electron II has $$n = 3$$ while electron III has $$n = 4$$. Because the smaller $$n$$ corresponds to lower energy, electron II precedes electron III.

Electron I has the largest value, $$n + l = 6$$, so it is the highest in energy.

Collecting all these comparisons we obtain the increasing energy order:

$$\text{IV} \;<\; \text{II} \;<\; \text{III} \;<\; \text{I}.$$

This sequence matches exactly with Option B.

Hence, the correct answer is Option B.