Assign the position of the element having outer electronic configuration in the periodic table:
$(i)$ $ns^{2} np^{4}$ for $n=3$
$(ii)$ $(n-1)d^{2} ns^{2}$ for $n=4$
$(iii)$ $(n-2)f^{7} (n-1)d^{1} ns^{2}$ for $n=6$

(N/A) $(i)$ Since $n=3$,the element belongs to the $3^{\text{rd}}$ period. It is a $p$-block element since the last electron occupies the $p$-orbital.
There are $4$ electrons in the $p$-orbital. Thus,the group number $= 2 (s\text{-block}) + 10 (d\text{-block}) + 4 (p\text{-electrons}) = 16$.
Therefore,the element belongs to the $3^{\text{rd}}$ period and $16^{\text{th}}$ group. The element is Sulphur $(S)$.
$(ii)$ Since $n=4$,the element belongs to the $4^{\text{th}}$ period. It is a $d$-block element as $d$-orbitals are being filled.
There are $2$ electrons in the $d$-orbital. Thus,the group number $= 2 (s\text{-block}) + 2 (d\text{-electrons}) = 4$.
Therefore,it is a $4^{\text{th}}$ period and $4^{\text{th}}$ group element. The element is Titanium $(Ti)$.
$(iii)$ Since $n=6$,the element is in the $6^{\text{th}}$ period. It is an $f$-block element (Lanthanoid series). All $f$-block elements belong to group $3$.
Its electronic configuration is $[Xe] 4f^{7} 5d^{1} 6s^{2}$. The atomic number is $54 + 7 + 1 + 2 = 64$. The element is Gadolinium $(Gd)$.