If the number of terms in the expansion of ${\left(1+5x+10x^2+10x^3+5x^4+x^5\right)}^{20}$ is $m,$ then unit’s place of $2^m$ is

$\begin{array}{l}{\left(1+5x+10x^2+10x^3+5x^4+x^5\right)}^{20}\\ ={\left((1+x{)}^5\right)}^{20}=(1+x{)}^{100}\end{array}$

$m$ = number of terms = 101,

We have

$\begin{array}{l}{2}^m=2^{101}={\left(2^5\right)}^{20}\\ =(30+2{)}^{20}(2)=\left(10m_1+2^{20}\right)\end{array}$

Let $k$ = Unit's digit of $2^m$ is

= units digit of $2^{21}$

Now, $2^{21}={\left(2^5\right)}^4\left(2\right)=(30+2{)}^4(2)$

$\therefore$ k = unit’s digit of $2^5=2.$