It is known that $(r+1)^{th}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of $(a+b)^{n}$ is given by

${T_{r + 1}} = {\,^n}{C_r}{a^{n – r}}{b^r}$

Assuming that $x^{2}$ occurs in the $(r+1)^{\text {th }}$ term of the expansion of $(1+x)^{m}$, we obtain

${T_{r + 1}} = {\,^m}{C_r}{(1)^{m – r}}{(x)^r} = {\,^m}{C_r}{(x)^r}$

Comparing the indices of $x$ in $x^{2}$ and in $T_{r+1},$ we obtain $r=2$

Therefore, the coefficient of $x^{2}$ is $^{m} C_{2}$

It is given that the coefficient of $x^{2}$ in the expansion $(1+x)^{m}$ is $6$

$\therefore {\,^m}{C_2} = 6$

$\Rightarrow \frac{m !}{2 !(m-2) !}=6$

$\Rightarrow \frac{m(m-1)(m-2) !}{2 \times(m-2) !}=6$

$\Rightarrow m(m-1)=12$

$\Rightarrow m^{2}-m-12=0$

$\Rightarrow m^{2}-4 m+3 m-12=0$

$\Rightarrow m(m-4)+3(m-4)=0$

$\Rightarrow(m-4)(m+3)=0$

$\Rightarrow(m-4)=0$ or $(m+3)=0$

$\Rightarrow m=4$ or $m=-3$

Thus, the positive value of $m$, for which the coefficient of $x^{2}$ in the expansion $(1+x)^{m}$ is $6.$ is $4$